Explore topic-wise InterviewSolutions in Class 11.

This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your Class 11 knowledge and support exam preparation. Choose a topic below to get started.

49501.

A sample of oxygen gas and a sample of an unknown gas are weighed separately in the same evacuated flask. Use thedata given below to find the molar mass of the unknown gas (assume experiments are carried out at the same pressure and temperature). |:(" Mass of evacuted flask",,124.46 g),(" Mass of flask+oxygen",,125.10g),("Mass of flask+unknown gas",,125.34 g):|

Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/22-294057" style="font-weight:bold;" target="_blank" title="Click to know more about 22">22</a> g/mol<br/>38g/mol<br/>44g/mol<br/>84g/mol</p>Answer :<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a></body></html>
Discussion
49502.

A sample of optically active alochol has a specific ratation, [alpha]_(D)^(25), equalto +1.51^(@). a. What is the percentage enantiomeric excess of the sample? b. Which enantiomer is in excess, (R ) or (S)?

Answer» <html><body><p></p>Solution :a. Enantiomeric excess `= ("<a href="https://interviewquestions.tuteehub.com/tag/observed-582965" style="font-weight:bold;" target="_blank" title="Click to know more about OBSERVED">OBSERVED</a> rotation")/("Specific rotation of <a href="https://interviewquestions.tuteehub.com/tag/pure-1172806" style="font-weight:bold;" target="_blank" title="Click to know more about PURE">PURE</a> enantiomer")<a href="https://interviewquestions.tuteehub.com/tag/xx100-3292680" style="font-weight:bold;" target="_blank" title="Click to know more about XX100">XX100</a>` <br/> `=(+1.151^(@))/(+5.756^(@))xx100= 20.00%` <br/> <a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a>. Since the (<a href="https://interviewquestions.tuteehub.com/tag/r-611811" style="font-weight:bold;" target="_blank" title="Click to know more about R">R</a> ) enantiomer is `(+)`-from, it is present in excess</body></html>
Discussion
49503.

A sample of oleum is such that ratio of free SO_(3) by combined SO_(3) is equal to unity. Calculate its labelling in terms of percentage oleum.

Answer» <html><body><p><br/></p>Solution :Let <a href="https://interviewquestions.tuteehub.com/tag/free-465311" style="font-weight:bold;" target="_blank" title="Click to know more about FREE">FREE</a> `SO_(3)<a href="https://interviewquestions.tuteehub.com/tag/rarr-1175461" style="font-weight:bold;" target="_blank" title="Click to know more about RARR">RARR</a> xg` <br/> `SO_(3)` in form of `H_(2)SO_(4)` <br/> `rarr(x)/(<a href="https://interviewquestions.tuteehub.com/tag/80-337972" style="font-weight:bold;" target="_blank" title="Click to know more about 80">80</a>)xx98=1.225x` <br/> so total<br/> `x+1.225x=100` <br/> `x=449.49` <br/> water required `=(44.94)/(80) xx18 = 10..11 <a href="https://interviewquestions.tuteehub.com/tag/g-1003017" style="font-weight:bold;" target="_blank" title="Click to know more about G">G</a>%` oleum `=100+10.11=110.11%` <br/></body></html>
Discussion
49504.

A sample of oxygen consists of three isotopes O - 16, O - 17 and O - 18 in the relative abundance of 99.76 %, 0.04 % and 0.20 % respectively. What is the average atomic mass of oxygen ?

Answer» <html><body><p><br/></p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/according-366619" style="font-weight:bold;" target="_blank" title="Click to know more about ACCORDING">ACCORDING</a> to available data : <br/> <a href="https://interviewquestions.tuteehub.com/tag/average-13416" style="font-weight:bold;" target="_blank" title="Click to know more about AVERAGE">AVERAGE</a> <a href="https://interviewquestions.tuteehub.com/tag/atomic-2477" style="font-weight:bold;" target="_blank" title="Click to know more about ATOMIC">ATOMIC</a> mass of oxygen `= 16 <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> 0.9976 + 17 xx0.0004 + <a href="https://interviewquestions.tuteehub.com/tag/18-278913" style="font-weight:bold;" target="_blank" title="Click to know more about 18">18</a> xx 0.002` <br/> `= 15.9616 + 0.0068 + 0.036 = 16.004 u`.</body></html>
Discussion
49505.

A sample of nitrogen gas occupies a volume of 320cm^(3) at STP. Calculate its volume at 66^(@)C and 0*825 bar presssure.

Answer» <html><body><p></p>Solution :Here , `P_(1)1*<a href="https://interviewquestions.tuteehub.com/tag/00-254995" style="font-weight:bold;" target="_blank" title="Click to know more about 00">00</a> "<a href="https://interviewquestions.tuteehub.com/tag/bar-892478" style="font-weight:bold;" target="_blank" title="Click to know more about BAR">BAR</a>", P_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>) 0*825`bar <br/> `V_(1)=320 cm^(3) "" V_(2)=?` <br/> `T_(1)=273*15K "" T_(2)66^(@)C=339*15 K` <br/> According to the <a href="https://interviewquestions.tuteehub.com/tag/gas-1003521" style="font-weight:bold;" target="_blank" title="Click to know more about GAS">GAS</a> equation, <br/> `(P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2))=(1xx320xx339)/(273xx0*825)=481*6 cm^(3)` <br/>`v_(2)=481*6 cm^(3)`</body></html>
Discussion
49506.

A sample of nitrogen gas occupies a volume of 1 L at a pressure of 0.5 bar at 40^@C. Calculate the pressure if the gas is compressed to 0.225 cm^3"at" -6^@C

Answer» <html><body><p></p>Solution :In the present case, <br/> `P_1 = 0.500` atm, <br/> `V_1 = 1.000 <a href="https://interviewquestions.tuteehub.com/tag/l-535906" style="font-weight:bold;" target="_blank" title="Click to know more about L">L</a>`, <br/> `T_1 = 273 + <a href="https://interviewquestions.tuteehub.com/tag/40-314386" style="font-weight:bold;" target="_blank" title="Click to know more about 40">40</a> = 313 K " and " P_2` =?, <br/> `V_2 = 0.225 cm^3 =(0.225)/(<a href="https://interviewquestions.tuteehub.com/tag/1000-265236" style="font-weight:bold;" target="_blank" title="Click to know more about 1000">1000</a>)L`, <br/>`T_2 = -6+ 273 = 267 K` <br/> ( `:.10^3 cm^3 = 1 L)` <br/>According to the gas <a href="https://interviewquestions.tuteehub.com/tag/equation-974081" style="font-weight:bold;" target="_blank" title="Click to know more about EQUATION">EQUATION</a>, `(P_1 V_1)/T_1 = (P_2 V_2)/T_2`<br/> On substituting the values, we have <br/> `(0.500 xx 1.000)/313 = (p_2 xx (0.225)/1000)/267` <br/> `:. "" P_2 = 1.89xx10^3` atm <br/> Therefore, the <a href="https://interviewquestions.tuteehub.com/tag/given-473447" style="font-weight:bold;" target="_blank" title="Click to know more about GIVEN">GIVEN</a> gas should be compressed to a pressure of `1.89 xx 10^3` atm.</body></html>
Discussion
49507.

A sample of nitric acid is 55 per cent by mass. Calculate the mass of nitric acid present in 100 cm of the sample if its density is 1.36 g cm^(-3).

Answer» <html><body><p></p>Solution :Since, the given sample is 55% by mass, 100 g of the sample would <a href="https://interviewquestions.tuteehub.com/tag/contain-409810" style="font-weight:bold;" target="_blank" title="Click to know more about CONTAIN">CONTAIN</a> 55 g of `HNO_3`. <br/> Since, <a href="https://interviewquestions.tuteehub.com/tag/density-17451" style="font-weight:bold;" target="_blank" title="Click to know more about DENSITY">DENSITY</a> `=("mass")/("volume")` <br/> Therefore, the volume of 100 g of the sample `=("mass")/("density") = 100/(1.36) = 73.5 cm^(3)` <br/> Thus,73.5 cm of the given sample contains 55 g of `HNO_3`. <br/> Hence, the mass of `HNO_3` present in 100 cm of the sample <br/> `=55/(73.5) <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> 100 = 74.8 g`</body></html>
Discussion
49508.

A sampleof NaOH (s) was added to water in a calorimeter. The temperature was monitoredas the NaOH dissolved to give the data below . Determinethe heat released during the solution process.(Assume the solution specific heat is 4.18J.g^(-1).K^(-1))

Answer» <html><body><p>`1.01xx10^(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)` <a href="https://interviewquestions.tuteehub.com/tag/joules-1062246" style="font-weight:bold;" target="_blank" title="Click to know more about JOULES">JOULES</a><br/>`2.66xx10^(3)`Joules<br/>`1.01xx10^(<a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>)`Joules<br/>`1.11xx10^(4)`Joules</p>Answer :d</body></html>
Discussion
49509.

A sample of Na_2CO_3H_2O weighing 0.62 g is added to 100 mL of 0.1 N H_2SO_4. Will the resulting solution be acidic, basic or neutral ?

Answer» <html><body><p><br/></p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :<a href="https://interviewquestions.tuteehub.com/tag/neutral-1114358" style="font-weight:bold;" target="_blank" title="Click to know more about NEUTRAL">NEUTRAL</a></body></html>
Discussion
49510.

A sample of N_(2)O_(4(g)) with a pressure of I atm is placed in a flask. When equilibrium is reached, 20% N_(2)O_(4(g)) has been converted to NO_(2(g)) .N_(2)O_(4(g)) hArr 2NO_(2(g)), If the orginal pressure is made 10% of the earlier, then percent dissociation will be

Answer» <html><body><p>0.<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a><br/>0.42<br/>0.54<br/>0.62</p>Solution :<img src="https://doubtnut-static.s.llnwi.net/static/physics_images/AKS_ELT_AO_CHE_XI_V01_B_C04_E03_006_S01.png" width="80%"/> <br/> Parpal pressure `((1-x)/(1+x))P, ((2x)/(1+x))P` <br/> `K_(P)=(4x^(2)P)/(1-x^(2))` <br/> In case-I: P=1 atm and x = 0.20 <br/> In case II : P=0.1 atm 10% of the <a href="https://interviewquestions.tuteehub.com/tag/earlier-963550" style="font-weight:bold;" target="_blank" title="Click to know more about EARLIER">EARLIER</a> pressure and x=? <br/> `(4(0.2)^(2) <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> 1)/([1-(0.2)^(2)])=(4x^(2)(0.1))/(1-x^(2))` <br/> `(0.04)/(1-0.04)=(x^(2)(0.1))/((1-x^(2)))(x^(2))/(1-x^(2))=0.4167` <br/> `x^(2)=0.2941, x=0.2941 <a href="https://interviewquestions.tuteehub.com/tag/implies-1037962" style="font-weight:bold;" target="_blank" title="Click to know more about IMPLIES">IMPLIES</a>` <a href="https://interviewquestions.tuteehub.com/tag/thus-2307358" style="font-weight:bold;" target="_blank" title="Click to know more about THUS">THUS</a> 54.23%</body></html>
Discussion
49511.

A sample of municipal water contains one part of urea (molecular wt = 60) per million parts of water by weight. The number of urea molecules in a drop of water of volume 0.05ml is

Answer» <html><body><p>`2.5xx10^(14)`<br/>`5xx10^(14)`<br/>`5xx10^(<a href="https://interviewquestions.tuteehub.com/tag/13-271882" style="font-weight:bold;" target="_blank" title="Click to know more about 13">13</a>)`<br/>`5xx10^(15)`</p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :B</body></html>
Discussion
49512.

A sample of mixed alkalis containing NaOH and Na_(2)CO_(3) is titrated in the following two schemes : (i) 10 ml of above mixture requires 8 ml of 0.1 N HCl by using phenolphthalein. (ii) 10 ml of above mixture requires 10 ml of 0.1 N HCl by using methyl orange. Calculate the ratio of the weight of NaOH and Na_(2)CO_(3) in the sample mixture.

Answer» <html><body><p></p>Solution :Basic principle involved is as follows : <br/> Acid used with <a href="https://interviewquestions.tuteehub.com/tag/phenolphthalein-599562" style="font-weight:bold;" target="_blank" title="Click to know more about PHENOLPHTHALEIN">PHENOLPHTHALEIN</a> as <a href="https://interviewquestions.tuteehub.com/tag/indicator-1041513" style="font-weight:bold;" target="_blank" title="Click to know more about INDICATOR">INDICATOR</a> = Complete neutralization ofNaOH `+ (1)/(2)` neutralization of `Na_(2)CO_(3)` <br/> Acid used with methyl orange as indicator= Complete neutralization of NaOH + complete neutralization of `Na_(2)CO_(3)` <br/> <a href="https://interviewquestions.tuteehub.com/tag/meq-1093952" style="font-weight:bold;" target="_blank" title="Click to know more about MEQ">MEQ</a> of HCl used for 10 ml of mixture using phenolphthalein `= 8 <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> 0.1 = 0.8` <br/> `:. ` meq of`NaOH + (1)/(2) M "eq of"Na_(2)CO_(3) = 0.8`...(i)<br/> meq of HCl usedfor 10 ml of mixture using methly orange `= 10 xx 0.1 = 1 ` <br/> `:.` meq of NaOH + M eq of `Na_(2)CO_(3)=1` <br/> From eqns (i) and (<a href="https://interviewquestions.tuteehub.com/tag/ii-1036832" style="font-weight:bold;" target="_blank" title="Click to know more about II">II</a>), meq of `Na_(2)CO_(3) = (1-0.8)xx2=0.4 ,` meq of `NaOH = 1 - 0.4 = 0.6` <br/> Eq. wt of `Na_(2)CO_(3) = 106//2=53`, Eq. wt of NaOH = 40 <br/> `:. ` 0.4 meq of `Na_(2)CO_(3) = (53)/(1000) xx0.4 = 0.0212 g ` <br/> 0.6meq of `NaOH = (40)/(1000) xx 0.6 = 0.024 g` <br/> Ratio of weight of NaOH and `Na_(2)CO_(3)=0.024//0.0212 = 1.132`</body></html>
Discussion
49513.

A sample of Mg metal containing some MgO as impurity was dissolved in 125 mL of 0.1N H_(2)SO_(4). The volume of H_(2) evolved at 27.5^(circ)C and 1 atmwas 120.0 mL. The resulting solution was found to be 0.02N with respect to H_(2)SO_(4). Calculate the weight of sample dissolved and the % by weight of pure Mg metal in sample. Neglect any change in volume.

Answer» <html><body><p><br/></p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :`122.16 <a href="https://interviewquestions.tuteehub.com/tag/g-1003017" style="font-weight:bold;" target="_blank" title="Click to know more about G">G</a>, Mg=95.57%`;</body></html>
Discussion
49514.

A sample of magnesium is partially oxidised to magnesia. 3 grams of such sample is treated with excess dilute sulphuric acid and the hydrogen collected measures 1.12 L at STP. What is the weight ratio of metal and metal oxide in the sample?

Answer» <html><body><p></p>Solution :Reaction of sulphuric <a href="https://interviewquestions.tuteehub.com/tag/acid-847491" style="font-weight:bold;" target="_blank" title="Click to know more about ACID">ACID</a> with <a href="https://interviewquestions.tuteehub.com/tag/components-926700" style="font-weight:bold;" target="_blank" title="Click to know more about COMPONENTS">COMPONENTS</a> is given as <br/> `<a href="https://interviewquestions.tuteehub.com/tag/mg-1095425" style="font-weight:bold;" target="_blank" title="Click to know more about MG">MG</a>+H_(2)SO_(4) to MgSO_(4)+H_(2)` <br/> `MgO+H_(2)SO_(4) to MgSO_(4)+H_(2)O` <br/> 1 mole of `H_(2)=` 1 mole of Mg <br/> 22.4 of `H_(2)` at STP=24 g of Mg <br/> 1.12 L of `H_(2)` at =STP=? <br/> Weight of magnesium in given sample= <br/> `(1.12)/(22.4) <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> 24=1.2" grams"` <br/> Weight of MgO in given sample `=3-1.2=1.8` grams <br/> Weight ratio of Mg and MgO =1.2 : 1.8=2:3</body></html>
Discussion
49515.

The common name of calcium sulphate hemihydrate is-

Answer» <html><body><p><br/></p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :3.8g</body></html>
Discussion
49516.

A sample of KCl is contaminated with NaCl. 4.176 g of the sample is dissolved in distilled water and the solution is made to 500 mL . 25 mL of the above solution required 27.50 mL of a solution of silver nitrate (normality factor 0.115) to react completely with it. Calculate the percentage contamination of the sample.

Answer» <html><body><p><br/></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :N//A</body></html>
Discussion
49517.

A sample of hydrogen peroxide is labelled as 10 volume. Its strength is g L^(-1) is _________.

Answer» <html><body><p>`30.00`<br/>`60.70`<br/>`15.17`<br/>`45.42`</p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :A</body></html>
Discussion
49518.

A sample of hydrochloric acid contains 20 % hydrochloric acid by weight. How much of this acid is required for the complete reduction of 50 g of calcium carbonate ?

Answer» <html><body><p><br/></p>Solution :Chemical equation for the reaciton is : <br/> `underset(underset(=100g)(40+12+48))(CaCO_(3))+underset(underset(=73g)(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>(1+35.5)))(2HCl)rarrCaCl_(2)+H_(2)O+CO_(2)` <br/> <a href="https://interviewquestions.tuteehub.com/tag/100-263808" style="font-weight:bold;" target="_blank" title="Click to know more about 100">100</a> g of calcium carbonate require acid = <a href="https://interviewquestions.tuteehub.com/tag/73-334616" style="font-weight:bold;" target="_blank" title="Click to know more about 73">73</a> g <br/> 50 g of calcium carbonate require acid `= (73)/(100) xx 50 = 36.5 g` <br/> Mass of the <a href="https://interviewquestions.tuteehub.com/tag/sample-1194587" style="font-weight:bold;" target="_blank" title="Click to know more about SAMPLE">SAMPLE</a> of hydrochloric `= 36.5 xx (100)/(20) = 182.5 g`.</body></html>
Discussion
49519.

A sample of hydrated copper sulphate is heated to drive off the water of crystallization, cooled and reweighed 0.869 g of CuSO_(4)aH_(2)O gave a residue of 0.556 g. Find the molecular formula of hydrated copper sulphate.

Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :0.869 g of `CuSO_(4)*aH_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)O` gave a residue of 0.556 g of Anhydrous `CuSO_(4)` `:` Weight of a `H_(2)O` molecule = 0.869-0.556 = 0.313 g<br/>Molecular weight of `H_(2)O = (1xx2)+16=2+16=18`<br/>No. of moles of water =`("Mass")/("Molecular mass")`<br/>`CuSO_(4)*5H_(2)O` - Molecular mass = 63.5 + <a href="https://interviewquestions.tuteehub.com/tag/32-307188" style="font-weight:bold;" target="_blank" title="Click to know more about 32">32</a> + 64 + <a href="https://interviewquestions.tuteehub.com/tag/90-341351" style="font-weight:bold;" target="_blank" title="Click to know more about 90">90</a> = 249.5 g<br/>249.5 gof `CuSO_(4) .5H_(2)O` on heating gives 159.5 g of `CuSO_(4)`<br/>. 0.869 g of `CuSO_(4)*aH_(2)O` on heating gives<br/>`159.5/249.5 xx 0.869`<br/>= 0.556 g of anhydrous `CuSO_(4)`<br/>`:.` a =5<br/>The molecular formula of hydrated copper <a href="https://interviewquestions.tuteehub.com/tag/sulphate-1234306" style="font-weight:bold;" target="_blank" title="Click to know more about SULPHATE">SULPHATE</a> = `CuSO_(4)*5H_(2)O.`</body></html>
Discussion
49520.

A sample of HI_((g))is placed in flask at a pressure of 0.2 atm. At equilibrium the partial pressure of HI_((g))is 0.04 atm. What is K_p for thegiven equilibrium ? 2HI_((g)) hArr H_(2(g)) + I_(2(g))

Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :`{:("Reaction:", 2HI_((g)) hArr, H_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>(g)) + , I_(2(g))),("Initial <a href="https://interviewquestions.tuteehub.com/tag/pressure-1164240" style="font-weight:bold;" target="_blank" title="Click to know more about PRESSURE">PRESSURE</a>:","0.2 atm",zero,zero),("Change in reaction :", "-2x atm","x atm","x atm"),("Partial pressure at <a href="https://interviewquestions.tuteehub.com/tag/equilibrium-974342" style="font-weight:bold;" target="_blank" title="Click to know more about EQUILIBRIUM">EQUILIBRIUM</a> :", (0.2-2x), "x atm","x atm"),(,=0.04 "atm", =0.08 "atm" , =0.08 "atm"):}` <br/> At equilibrium partial pressure of <a href="https://interviewquestions.tuteehub.com/tag/hi-479908" style="font-weight:bold;" target="_blank" title="Click to know more about HI">HI</a> = 0.2 - 2x = 0.04<br/> `therefore` 2x=0.2 - 0.04 =0.16 <br/> `therefore` x=0.08 atm <br/> Thus, `p_(HI)`= 0.04 , `p_(H_2)` = 0.08 = `p_(I_2)`atm <br/> `K_p=((p_(H_2))(p_(I_2)))/((p_(HI))^2)="(0.08 atm)(0.08 atm)"/"(0.04 atm)"^2=4.0`</body></html>
Discussion
49521.

A sample of HI (g) is placed in a flask at a pressure of 0*2 atm. At equilibrium , the partial pressure of HI (g) is 0*04 atm . What is K_(p) for the given equilibrium ?

Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :` {:(,2 <a href="https://interviewquestions.tuteehub.com/tag/hi-479908" style="font-weight:bold;" target="_blank" title="Click to know more about HI">HI</a>(g),hArr,H_(2)(g),+,I_(2)), (" Intial pressure",0*2"atm",,0,,0), ("At <a href="https://interviewquestions.tuteehub.com/tag/eqm-446398" style="font-weight:bold;" target="_blank" title="Click to know more about EQM">EQM</a>",0*04,,(0*16)/2,,(0*16)/2),(,,,=0*8"atm",,=0*8 "atm"),(,,,,,("Decease in the pressure <a href="https://interviewquestions.tuteehub.com/tag/pf-1152667" style="font-weight:bold;" target="_blank" title="Click to know more about PF">PF</a> Hi" = 0*2 - 0*04 = 0*16 "atm")):}`<br/> ` :. K_(p) = ( pH_(2) xx pI_(2))/(p_(HI_(2))) = ( 0*8 "atm " xx 0*08 "atm")/(0*04 "atm")^(2)= 4*0 . `</body></html>
Discussion
49522.

A sample of HI(g) is placed in a flask a pressure of 0.2 atm. At equilibrium partial pressure of HI(g) is 0.04atm. What is K_p for the given equilibrium ? 2HI(g) hArr H_(2)(g) + I_(2)(g)

Answer» <html><body><p></p>Solution :`<a href="https://interviewquestions.tuteehub.com/tag/phi-599602" style="font-weight:bold;" target="_blank" title="Click to know more about PHI">PHI</a> =0.04` <a href="https://interviewquestions.tuteehub.com/tag/atm-364409" style="font-weight:bold;" target="_blank" title="Click to know more about ATM">ATM</a> `pH_2 = 0.08` atm `Pl_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)` =0.08 atm <br/> `K_(p) = (PH_2 xxpl_2)/(P_(HI)^2) = ((0.08 atm )xx(0.08))/((0.04 atm ) xx (0.04 atm))=4.0`</body></html>
Discussion
49523.

A sample of helium has a volume of500 cm^3at 373 K. Calculate the temperature at which the volume will become 260 cm^3 . Assume that the pressure is constant.

Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :Given that, <br/> `V_1 = 500 cm^3 , ""V_2 = 260 cm^3`, <br/> `T_1 = 373 <a href="https://interviewquestions.tuteehub.com/tag/k-527196" style="font-weight:bold;" target="_blank" title="Click to know more about K">K</a>, "" T_2` = ? <br/> Since pressure remains <a href="https://interviewquestions.tuteehub.com/tag/constant-930172" style="font-weight:bold;" target="_blank" title="Click to know more about CONSTANT">CONSTANT</a>, hence, according to Charles. law, we have <br/> `V_1/T_1 = V_2/T_2` <br/> or `T_2 = (V_2 T_1)/(V_1) =(260xx 373)/(500) = 194 K` <br/>Hence, the required <a href="https://interviewquestions.tuteehub.com/tag/temperature-11887" style="font-weight:bold;" target="_blank" title="Click to know more about TEMPERATURE">TEMPERATURE</a> is 194 K.</body></html>
Discussion
49524.

A sample of hard water contains 96 ppm of SO_(4)^(2-) and 183 ppm of HCO_(3)^(-)"with" Ca^(2+) as the only cation. How many moles of CaO will be required to remove HCO_(3)^(-) from 1000 kg of this water ? If 1000 kg of this water is treated with the amount of CaO calculated above, what will be the concentration (in ppm) of residual Ca^(2+) ions ? (Assume CaCO_(3) to becompletely insoluble in water). If the Ca^(2+) ions in one litre of the treated water are completely exchanged with hydrogen ions, what will be its pH ? (One ppm means one part of the substance in one million parts of water , weight/weight).

Answer» <html><body><p></p>Solution :`SO_(4)^(2-)` present in 1000 kgof water `=(96)/(10^(6))xx1000 kg = 96 g = (96)/(96) =1 ` mole <br/> `HCO_(3)^(-)` present in 1000 kg of water `= (183)/(10^(6))xx1000 kg = 183 g = (183)/(61)=3 ` moles <br/> `Ca^(2+)` present along with `SO_(4)^(2-)` ions = 1 mole <br/> `Ca^(2+)` present along with `HCO_(3)^(-) ` as `Ca(HCO_(3))_(2) = (3)/(2) ` mole <br/> `:.` Total `Ca^(2+)` present in 1000 kg of water ` = 1 +(3)/(2) = 2.5` moles <br/> CaO added will react with `Ca(HCO_(3))_(2)` as follows : <br/> `Ca(HCO_(3))_(2) + CaO rarr 2 CaCO_(3) harr + H_(2)O` <br/> But `Ca(HCO_(3))_(2)` present `= (3)/(2)` mole (calculated above) <br/> `:.` CaO required `= (3)/(2) ` mole = 1.5 moles <br/> After treatment with CaO i.e., removal of `Ca(HCO_(3))_(2)`, amount of `Ca^(2+) ` <a href="https://interviewquestions.tuteehub.com/tag/left-1070879" style="font-weight:bold;" target="_blank" title="Click to know more about LEFT">LEFT</a> (<a href="https://interviewquestions.tuteehub.com/tag/due-433472" style="font-weight:bold;" target="_blank" title="Click to know more about DUE">DUE</a> to `CaSO_(4)` only ) in 1000 kgof water = 1 mole = 40 g <br/> `:.` Concentration of residual `Ca^(2+)` (in ppm) = 40 ppm <br/> Now 1000 kg of water contain `Ca^(2+) = 10^(-3)` mole<br/> No. of mole of `H^(+)` <a href="https://interviewquestions.tuteehub.com/tag/exchanged-2075520" style="font-weight:bold;" target="_blank" title="Click to know more about EXCHANGED">EXCHANGED</a> `= 2 xx 10^(-3)` mole<br/> `:. pH= - log (2xx10^(-3))= 2.7`</body></html>
Discussion
49525.

A sample of hard water is allowed to pass through an anion exchanger. Will it produce lather will soap easily?

Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :No. `<a href="https://interviewquestions.tuteehub.com/tag/ca-375" style="font-weight:bold;" target="_blank" title="Click to know more about CA">CA</a>^(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>+) and Mg^(2+)` ions are still present and these will react with <a href="https://interviewquestions.tuteehub.com/tag/soap-11667" style="font-weight:bold;" target="_blank" title="Click to know more about SOAP">SOAP</a> to form curdy white ppt. Therefore, it will not preduce lather with soap easily.</body></html>
Discussion
49526.

A sample of hard water contains 1 xx 10^(-3) mole of CaCl_2per 10kg of water. Calculate the degree of hardness of water sample

Answer» <html><body><p></p>Solution :Weight of `CaCl_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)` present in 10kg of <a href="https://interviewquestions.tuteehub.com/tag/water-1449333" style="font-weight:bold;" target="_blank" title="Click to know more about WATER">WATER</a> =`<a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a>^(3)` moles = <a href="https://interviewquestions.tuteehub.com/tag/111-268410" style="font-weight:bold;" target="_blank" title="Click to know more about 111">111</a>`xx10^(-3)g` <br/> Weight of `CaCl_(2)` present in `10^(6)` of water `=(111xx10^(-3)xx10^(6))/(10(4))=11.1g` <br/> 111g of `CaCl_(2)-=100g ` of `CaCO_(3),11.1g` of `CaCl-=10g` of `CaCO_(3)` <br/> The degree of hardness of given sample=10ppm</body></html>
Discussion
49527.

A sample of hard water contains 100 ppm of CaSO_(4).What minimum fraction of water shouldbe evaporated off so that solid CaSO_(4) begins to separate out ? K_(sp)for CaSO_(4)is 9.0xx10^(-6).

Answer» <html><body><p></p>Solution :Maximum solubility of `CaSO_(4)` in water can be calculated from its `K_(sp)` value as <a href="https://interviewquestions.tuteehub.com/tag/follows-994526" style="font-weight:bold;" target="_blank" title="Click to know more about FOLLOWS">FOLLOWS</a>: <br/> `S=sqrt(K_(sp))=sqrt(9.0xx10^(-6))=3.0=10^s(-3) ` mol `<a href="https://interviewquestions.tuteehub.com/tag/l-535906" style="font-weight:bold;" target="_blank" title="Click to know more about L">L</a>^(-1)` <br/> Suppose the volume of water taken = V litre <br/> As `CaSO_(4)` present is 100 ppm, i.e., 100 g per `10^(6) g ` of water, therefore, `CaSO_(4)` present in V litres `(V xx 10^(3)g)` of water <br/> `=(100)/(10^(6))xxV xx 10^(3) g = 0.1 V g = (0.1 V)/(136)` moles(Molar mass of `CaSO_(4) = 136` g `"mol"^(-1)`) <br/> After evaporation, suppose volume of water left = V' litre <br/> <a href="https://interviewquestions.tuteehub.com/tag/thus-2307358" style="font-weight:bold;" target="_blank" title="Click to know more about THUS">THUS</a>, V' litre of water will now contain `= (0.1 V) /(136) ` moles of `CaSO_(4)`. <br/> This should be <a href="https://interviewquestions.tuteehub.com/tag/equal-446400" style="font-weight:bold;" target="_blank" title="Click to know more about EQUAL">EQUAL</a> to the maximum solubility in moles `L^(-1)`. <br/> `:. (0. V)/(136) xx (1)/(V')= 3.0 xx 10^(-3) or V' = (0.1 V)/(136 xx 3 xx 10^(-3))=0.245 V ` <br/> `:. ` Volume of water <a href="https://interviewquestions.tuteehub.com/tag/evaporated-7679270" style="font-weight:bold;" target="_blank" title="Click to know more about EVAPORATED">EVAPORATED</a>`= V - V' = V - 0.245 V = 0.755 V` <br/> i.e., 75.5 % of water should be evaporated off .</body></html>
Discussion
49528.

A sample of hard water contains 0.005 moles of calcium chloride per litre.What is the minimum concentration of sodium sulphate which must be added for removing theCa^(2+)ions from this water sample ?K_(sp) " of "CaSO_4= 2.4 xx 10^(-5)

Answer» <html><body><p>`4.8 xx <a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a>^(-2)`<br/>` 4.8 xx 10^(-3)` <br/>` 2.4 xx 10^(-2)` <br/>` 2.4 xx 10^(-3)` </p>Solution :`[<a href="https://interviewquestions.tuteehub.com/tag/ca-375" style="font-weight:bold;" target="_blank" title="Click to know more about CA">CA</a>^(+2)]= 5 xx 10 ^(-3)M` <br/>From`K_(<a href="https://interviewquestions.tuteehub.com/tag/sp-1219706" style="font-weight:bold;" target="_blank" title="Click to know more about SP">SP</a>)" of "CaSO_4` <br/>` 2.4 xx 10 ^(-5)= 5 xx 10^(-3)xx [SO_4 ^(-2) ]` <br/>` [SO_4^(-2)]= 4.8 xx 10 ^(-3)<a href="https://interviewquestions.tuteehub.com/tag/mz-1108227" style="font-weight:bold;" target="_blank" title="Click to know more about MZ">MZ</a>` <br/> ` [Na_2 SO_4 ] =4.8 xx 10^(-3)M`</body></html>
Discussion
49529.

A sample of H_(2)O_(2) is x % by mass. If x ml of KMnO_(4) are required to oxidize 1 gm of this H_(2)O_(2) sample, calculate the normality of KMnO_(4) solution.

Answer» <html><body><p>`0.46N`<br/>`0.5N`<br/>`0.6N`<br/>`0.65N`</p>Solution :Let mass of `H_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)O_(2)` solution = <a href="https://interviewquestions.tuteehub.com/tag/100-263808" style="font-weight:bold;" target="_blank" title="Click to know more about 100">100</a> gm <br/> mass of `H_(2)O_(2)` present is = <a href="https://interviewquestions.tuteehub.com/tag/xgm-3887063" style="font-weight:bold;" target="_blank" title="Click to know more about XGM">XGM</a> <br/> <a href="https://interviewquestions.tuteehub.com/tag/given-473447" style="font-weight:bold;" target="_blank" title="Click to know more about GIVEN">GIVEN</a> that mass of `H_(2)O_(2)` solution taken = <a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a> gm, <br/> So, mass of `H_(2)O_(2)` present in 1 gm solution <br/> `=(x)/(100)` Eq. of `H_(2)O_(2)=(x)/(100xx(34//2))` <br/> Eq. of `KMnO_(4)=NxxV(C)=Nxx x xx10^(-3)` <br/> Now `(x)/(100xx17)=Nxx x xx10^(-3)impliesN=0.59N`</body></html>
Discussion
49530.

A sample of gas occupies 12 L at 227^@C and at 1 atm pressure. The gas is cooled to - 73^@C at the same pressure. What would be the volume of the gas?

Answer» <html><body><p></p>Solution :Given that, <br/> `V_1 = 12 <a href="https://interviewquestions.tuteehub.com/tag/l-535906" style="font-weight:bold;" target="_blank" title="Click to know more about L">L</a>, "" V_2` =? <br/>`T_1 = 227 + <a href="https://interviewquestions.tuteehub.com/tag/273-1832932" style="font-weight:bold;" target="_blank" title="Click to know more about 273">273</a> = 500 K, " " T_2= -73 + 273 = 200 K` <br/> The <a href="https://interviewquestions.tuteehub.com/tag/pressure-1164240" style="font-weight:bold;" target="_blank" title="Click to know more about PRESSURE">PRESSURE</a> remains constant during cooling। Hence, <a href="https://interviewquestions.tuteehub.com/tag/according-366619" style="font-weight:bold;" target="_blank" title="Click to know more about ACCORDING">ACCORDING</a> to Charles. law, we have <br/> `V_1/T_1 = V_2/T_2`<br/>or `V_2 = (V_1 T_2)/(T_1) =(12xx 200)/(500) = 4.8 L`<br/>Hence, on cooling the volume of the given <a href="https://interviewquestions.tuteehub.com/tag/gas-1003521" style="font-weight:bold;" target="_blank" title="Click to know more about GAS">GAS</a> reduces to 4.8 L.</body></html>
Discussion
49531.

A sample of gas is compressed from an initial volume of 2V_(0) "to" V_(0) using three different processes. First: Using reversing isothermal Second : Using reversible adiabatic Third : Using irreversible adiabatic under a constant external pressure ,than :

Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/final-461168" style="font-weight:bold;" target="_blank" title="Click to know more about FINAL">FINAL</a> <a href="https://interviewquestions.tuteehub.com/tag/temerature-3101441" style="font-weight:bold;" target="_blank" title="Click to know more about TEMERATURE">TEMERATURE</a> of gas will be highest at the end of third process.<br/>Final temperature of gas will be highest at the end of second process <br/>Enthalpy <a href="https://interviewquestions.tuteehub.com/tag/change-913808" style="font-weight:bold;" target="_blank" title="Click to know more about CHANGE">CHANGE</a> of sample will be highest in isothermal process. (magnitude wise) <br/>Final pressure of gas will be highest at the end of second process.</p>Answer :A</body></html>
Discussion
49532.

A sample of gas has a volume of V_(1) litre at temperature t_(1).^(@)C. When the temperature of the gas is changed to t_(2).^(@)C at constant pressure, then the volume of the gas was found to increase by 10%. The percentage increase in temperature is

Answer» <html><body><p>0.<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a><br/>`(<a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a>+(2730)/(t_(1)))%`<br/>`<a href="https://interviewquestions.tuteehub.com/tag/20-287209" style="font-weight:bold;" target="_blank" title="Click to know more about 20">20</a>%`<br/>`(0.1+t_(1)^(-1))%`</p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a></body></html>
Discussion
49533.

A sample of gas has a volume of 0.2 lit measured at1 atm pressure and 0^@C. At the same pressure, but at 273^@C, its volume will become

Answer» <html><body><p>`0.1` <a href="https://interviewquestions.tuteehub.com/tag/lit-537267" style="font-weight:bold;" target="_blank" title="Click to know more about LIT">LIT</a><br/>`0.4 ` lit<br/>`0.8 ` lit<br/>0.6 lit</p>Answer :<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a></body></html>
Discussion
49534.

A sample of fuming sulphuric acid containing H_(2)SO_(4), SO_(3) and SO_(2) weighing 1.0 g is found to require 23. 47 mL of 1.0N alkali for its neutralisation. A separate sample shows the presence of 1.5% SO_(2). Find the percentage of free SO_(3),H_(2)SO_(4) and combined SO_(3) in the sample.

Answer» <html><body><p><br/></p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :`65.3%, 33.2%, 27.10%`;</body></html>
Discussion
49535.

A sample of Fe(SO_4)_3 and FeC_2O_4 was dissolved in H_2SO_4. 40 " mL of " (N)/(16)KMnO_4 was required for complete oxidation. After oxidation the mixture was reduced by (An)/(H_2SO_4). On again oxidation by same KMnO_4, 60 mL was required. Calculate the ratio of m" Eq of "Fe_2(SO_4)_3 and FeC_2O_4.

Answer» <html><body><p></p>Solution :`FeC_2O_4+H_2SO_4toFeSO_4+(COOH)_2` <br/> `Fe^(2+)toFe^(3+)+e^(-)` <br/> `C_2O_4^(2-)to2CO_2+2e^(-)` <br/> 40 " mL of " `(N)/(16)` `KMnO_4=(40)/(16)` m" Eq of "`(Fe^(+2)+OX^(2-))` <br/> Let us <a href="https://interviewquestions.tuteehub.com/tag/take-662846" style="font-weight:bold;" target="_blank" title="Click to know more about TAKE">TAKE</a> y m " mol of "`Fe^(2+)` <br/> `<a href="https://interviewquestions.tuteehub.com/tag/therefore-706901" style="font-weight:bold;" target="_blank" title="Click to know more about THEREFORE">THEREFORE</a> m" Eq of "Fe^(+2)=y` (n factor`=1`) <br/> y m " mol of "`OX^(2-)` <br/> `thereforem" Eq of "OX^(2-)=2y` (n factor`=2`) <br/> `thereforey+2y=(40)/(16)implies3y=(40)/(16)becausey=(40)/(48)` <br/> m" Eq of "`Fe^(2+)=(40)/(48)`, `m" Eq of "Fe_2C_2O_4=(40)/(16)` <br/> Second step: <br/> 60 " mL of " `(N)/(16)KMnO_4=(60)/(16)` m" Eq of "total `Fe^(2+)` <a href="https://interviewquestions.tuteehub.com/tag/ltbrgt-2804393" style="font-weight:bold;" target="_blank" title="Click to know more about LTBRGT">LTBRGT</a> `[Fe^(2+)` ions from `FeC_2O_4+Fe^(2+)` ions obtained after the reduction of `Fe_2(SO_4)_3]` <br/> m" Eq of "`Fe_2(SO_4)_3=(60)/(16)-(48)/(48)=(140)/(48)` <br/> <a href="https://interviewquestions.tuteehub.com/tag/ratio-13379" style="font-weight:bold;" target="_blank" title="Click to know more about RATIO">RATIO</a> of <a href="https://interviewquestions.tuteehub.com/tag/meq-1093952" style="font-weight:bold;" target="_blank" title="Click to know more about MEQ">MEQ</a>`(Fe_2(SO_4)_3)/(FeC_2O_4)=(140xx48)/(48xx40)=(7)/(6)=7:6`</body></html>
Discussion
49536.

A sample of ferrous sulphate and ferrous oxalate was dissolved in dil. H_(2)SO_(4) the complete oxidation of reactionmixture required 40 mL pf N//15 KMnO_(4). After the oxidation, the reactionmixture was reduced by Zn and H_(2)SO_(4). On again oxidation by same KMnO_(4), 25mL were required. Calucate the ratio of fe in ferrous sulphate and oxalate.

Answer» <html><body><p><br/></p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :`7//3`</body></html>
Discussion
49537.

A sample of ferrous oxide has actual formulaFe_(0.93) O_(1.00) . In this sample what fraction of metal ions are Fe^(2+) ions ? What type of non-stoichoimetric defect is present in the sample ?

Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :Metal deficency defect is present in the <a href="https://interviewquestions.tuteehub.com/tag/sample-1194587" style="font-weight:bold;" target="_blank" title="Click to know more about SAMPLE">SAMPLE</a> because <a href="https://interviewquestions.tuteehub.com/tag/iron-1051344" style="font-weight:bold;" target="_blank" title="Click to know more about IRON">IRON</a> is present in <a href="https://interviewquestions.tuteehub.com/tag/lesser-541161" style="font-weight:bold;" target="_blank" title="Click to know more about LESSER">LESSER</a> <a href="https://interviewquestions.tuteehub.com/tag/amount-374803" style="font-weight:bold;" target="_blank" title="Click to know more about AMOUNT">AMOUNT</a> than required according to stoichiometric composition .</body></html>
Discussion
49538.

A sample of ferrous oxide has actual formula Fe_0.93 O_1.00. In this sample what fraction of metal ions are Fe^(2+) ions ? What type of non-stoichoimetric defect is present in this sample ?

Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :<a href="https://interviewquestions.tuteehub.com/tag/metal-1094457" style="font-weight:bold;" target="_blank" title="Click to know more about METAL">METAL</a> deficiency defect is <a href="https://interviewquestions.tuteehub.com/tag/present-1163722" style="font-weight:bold;" target="_blank" title="Click to know more about PRESENT">PRESENT</a> in the sample because <a href="https://interviewquestions.tuteehub.com/tag/iron-1051344" style="font-weight:bold;" target="_blank" title="Click to know more about IRON">IRON</a> is present in lesser amount then required according to stoichiometric composition</body></html>
Discussion
49539.

A sample of drinking water was found to be severely contaminated with chloroform, CHCl_(3) supposed to be carcinogenic in nature. The level of contamination was 15 ppm ( by mass). (i) Express this in precent by mass. (ii) Determine the molality of chloroform in the water sample.

Answer» <html><body><p> <br/> <br/> </p>Solution :(i) 15 <a href="https://interviewquestions.tuteehub.com/tag/ppm-1162221" style="font-weight:bold;" target="_blank" title="Click to know more about PPM">PPM</a> means 15 parts of million, <br/> So, Mass percent `= (15xx100)/(10^(6))=1.5xx10^(-3)%` <br/> (ii) Molecular mass of `CHCl_(3)` <br/> `= 12.01+1.0079+(3xx35.45)` <br/> `= <a href="https://interviewquestions.tuteehub.com/tag/119-268764" style="font-weight:bold;" target="_blank" title="Click to know more about 119">119</a>.367 ~=119` gm/mol <br/> `1.5xx10^(-3)%` means, `1.5xx10^(-3)` gm chloroform in 100 gm <br/> Molality `= ("Weight of solute " xx 1000)/("Molecular mass of solute" xx "<a href="https://interviewquestions.tuteehub.com/tag/volume-728707" style="font-weight:bold;" target="_blank" title="Click to know more about VOLUME">VOLUME</a> of solution (in gm)")` <br/> `=(1.5xx10^(-3) xx 1000)/(119xx100)` <br/> `= 0.000126=1.26xx10^(-4)m`</body></html>
Discussion
49540.

A sample of ethane has the same mass as 10.0 million molecules of methane. How many C_2H_6 molecules does the sample contain?

Answer» <html><body><p></p>Solution :Let the number of `C_2H_6` molecules in the sample be n. As given, <a href="https://interviewquestions.tuteehub.com/tag/mass-1088425" style="font-weight:bold;" target="_blank" title="Click to know more about MASS">MASS</a> of `C_2H_6` = mass of `10^7` molecules of `CH_4` <br/> `n/("<a href="https://interviewquestions.tuteehub.com/tag/av-888878" style="font-weight:bold;" target="_blank" title="Click to know more about AV">AV</a>. <a href="https://interviewquestions.tuteehub.com/tag/const-11346" style="font-weight:bold;" target="_blank" title="Click to know more about CONST">CONST</a>") xx` mol wt. of `C_(2)H_(6) = 10^(7)/("Av. Const") xx` mol. Wt of `CH_(4)` <br/> `(n xx <a href="https://interviewquestions.tuteehub.com/tag/30-304807" style="font-weight:bold;" target="_blank" title="Click to know more about 30">30</a>)/("av. Const") = (10^(7) xx <a href="https://interviewquestions.tuteehub.com/tag/16-276476" style="font-weight:bold;" target="_blank" title="Click to know more about 16">16</a>)/("Av. const")` <br/> `therefore n = 5.34 xx 10^(6)`</body></html>
Discussion
49541.

A sample of drinking water was found to be severely contaminated with chloroform (CHCl _(3)) supposed to be a carcinogen. The level of contamination was 15 ppm (by mass) (i) express this in percentage by mass (ii) determine the molality of chloroform in the water sample.

Answer» <html><body><p></p>Solution :(i) <a href="https://interviewquestions.tuteehub.com/tag/15-274069" style="font-weight:bold;" target="_blank" title="Click to know more about 15">15</a> ppm means 15 parts in <a href="https://interviewquestions.tuteehub.com/tag/million-560845" style="font-weight:bold;" target="_blank" title="Click to know more about MILLION">MILLION</a> `(<a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a>^(6))` parts by mass in the solution <br/> `therefore %` of mass `= (15)/(10^(6) xx 100 = 1.5 xx 10 ^(-<a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>)` <br/> (ii) Taking 15 g chloroform in `10^(6)` g of the solution <br/> Mass of the solvent `= 10 ^(6)g` <br/> Molar mass of `CHCl_(3) = 12 + 1 + (3 xx 35.5) = 119.5 g mol^(-1)` <br/> `therefore`Molality `= ((15)/(119.5))/(10^(6)) xx 100 = 1.25 xx 10 ^(-5)m`</body></html>
Discussion
49542.

A sample of drinking water was found to be severely contaminated with chloroform, CHCl_(3), supposed to be carcinogenic in nature. The level of contamination was 15 ppm (by mass).(i) Express this in per cent by mass.(ii) Determine the molality of chloroform in the water sample.

Answer» <html><body><p></p>Solution :(i) 15 ppm (by mass) <a href="https://interviewquestions.tuteehub.com/tag/means-1091780" style="font-weight:bold;" target="_blank" title="Click to know more about MEANS">MEANS</a> 15 parts by mass of `CHCI_3` are present in `10^6` parts by mass of water. <br/> `therefore` Per cent by mass `=15/10^(6) <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> 100 = 1.5 xx 10^(-3) %` <br/> (ii) <a href="https://interviewquestions.tuteehub.com/tag/molar-562965" style="font-weight:bold;" target="_blank" title="Click to know more about MOLAR">MOLAR</a> mass of `CHCl_(3) = 12 +1 + (3 xx 35.5) = 119.5 g "mol"^(-1)` <br/> `therefore 10^(6)` g of <a href="https://interviewquestions.tuteehub.com/tag/sample-1194587" style="font-weight:bold;" target="_blank" title="Click to know more about SAMPLE">SAMPLE</a> contain chloroform = 15 g <br/> `therefore 10^(3)` g (1 kg) of sample will contain chloroform `=15/10^(6) xx 10^(3) = 1.5 xx 10^(-2) g` <br/> Number of moles of chloroform present in 1 kg of the sample <br/> `=(1.5 xx 10^(-2))/119.5` <br/> `=1.255 xx 10^(-4)` mole<br/> `therefore` Molality `=1.255 xx 10^(-4)` m</body></html>
Discussion
49543.

A sample of crystalline Ba(OH)_(2). XH_(2)O weight 1.578 gm was dissolved in water. The solution required 40 ml of 0.25 N HNO_(3) for complete relation. Determine the numberof molecular of water of crystallisat in base.

Answer» <html><body><p><br/></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :Eqts of `Ba(OH)_(2).xH_(2)O` = <a href="https://interviewquestions.tuteehub.com/tag/eq-446394" style="font-weight:bold;" target="_blank" title="Click to know more about EQ">EQ</a> of `HNO_(3)` <br/> `(1.578)/((171+18x)//2)=(<a href="https://interviewquestions.tuteehub.com/tag/40-314386" style="font-weight:bold;" target="_blank" title="Click to know more about 40">40</a>)/(1000)xx0.25` <br/> `18x=315.6-171impliesx=8`</body></html>
Discussion
49544.

A sample of CO with volume 500 mL at a pressure of 760 mm is to be compressed to a volume of 450 mL. What additional pressure is required to accomplish the change if the temperature is kept constant?

Answer» <html><body><p>&lt;<a href="https://interviewquestions.tuteehub.com/tag/p-588962" style="font-weight:bold;" target="_blank" title="Click to know more about P">P</a>&gt;</p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :84.4 mm <br/> <a href="https://interviewquestions.tuteehub.com/tag/using-7379753" style="font-weight:bold;" target="_blank" title="Click to know more about USING">USING</a> `P_(1)P_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)=P_(2)V_(2)`</body></html>
Discussion
49545.

A sample of coconut oil weighing 1.5763 g is mixed with 25 mL of 0.4210 M KOH. Some KOH is used in saponification of coconut oil. After the saponification is complete, 8.46 mL of 0.2732 M H_(2)SO_(4) is required to neutralize excess KOH. The saponification number of peanut oil is :

Answer» <html><body><p>209.6<br/>98.9<br/>108.9<br/>218.9</p>Solution :Number of <a href="https://interviewquestions.tuteehub.com/tag/milliequivalent-560828" style="font-weight:bold;" target="_blank" title="Click to know more about MILLIEQUIVALENT">MILLIEQUIVALENT</a> of KOH added`=25xx0.421=10.525` <br/> Number of milliequivalents <a href="https://interviewquestions.tuteehub.com/tag/left-1070879" style="font-weight:bold;" target="_blank" title="Click to know more about LEFT">LEFT</a> unreacted <br/> =Number of milliequivalents of `H_(2)SO_(4)` used <br/> `= 8.46xx0.2732xx24.623 "" ("Here, basicity of " H_(2)SO_(4)=2)` <br/> Number of milliequivalents of KOH used by oil =10.525-4.623=5.902 <br/> <a href="https://interviewquestions.tuteehub.com/tag/mass-1088425" style="font-weight:bold;" target="_blank" title="Click to know more about MASS">MASS</a> of KOH used `=(5.902xx56)/(<a href="https://interviewquestions.tuteehub.com/tag/1000-265236" style="font-weight:bold;" target="_blank" title="Click to know more about 1000">1000</a>)=0.3305 g =330.5 <a href="https://interviewquestions.tuteehub.com/tag/mg-1095425" style="font-weight:bold;" target="_blank" title="Click to know more about MG">MG</a>` <br/> Saponification number = Mass of KOH in mg used by 1 g oil or fat <br/> `=(0.3305xx1000)/(1.5763)=209.6`</body></html>
Discussion
49546.

A sample of CH_(4) of 0.08g was subjected to combustion at 27^(@)C in a bomb calorimeter. The temperature of the calorimeter system was found to be raised by 0.25^(@)C. If heat capacity of calorimeter is 18 kJ//""^(@)C, Delta H for combustion of CH_(4) at 27^(@)C is

Answer» <html><body><p>`-900KJ//"<a href="https://interviewquestions.tuteehub.com/tag/mole-1100299" style="font-weight:bold;" target="_blank" title="Click to know more about MOLE">MOLE</a>"`<br/>`-905KJ//"mole"`<br/>`-895KJ//"mole"`<br/>`-890KJ//"mole"`</p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a></body></html>
Discussion
49547.

A sample of calcium carbonate (CaCO_(3)) has the following percentage composition: Ca=40%,C=12%,O=48% If the law of constant proportions is true, then the weight off calcium in 4g of a sample of calcium carbonate from another source will be

Answer» <html><body><p>0.016 g<br/>0.16 g<br/>1.6 g<br/>16 g</p>Answer :<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a></body></html>
Discussion
49548.

A sample of calcium carbonate (CaCO_(3)) has the following percentage composition : Ca = 40 % C = 12 % , O = 48 % . If the law of constant proportions is true , then the weight of calcium in 4 g of a sample of calcium carbonate from another source will be

Answer» <html><body><p>0.016g<br/>0.16g<br/>1.6g<br/>16g</p>Answer :<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a></body></html>
Discussion
49549.

A sample of argon gas at 1 atm pressure and 27^(@)C expands reversibly and adiabatically from 1.25dm^(3) to 2.50 dm^(3). Calculatethe enthalpy changein this process. C_(v,m) for argonis 12.48 JK^(-1) mol^(-1)

Answer» <html><body><p></p>Solution :For adiabaticprocess , <br/> `T_(1)V_(1)^(gamma-1)= T_(2)V_(2)^(gamma-1) ` or `((V_(1))/(V_(2)))^(gamma-1) = (T_(2))/(T_(1))` <br/> But `gamma=( C_(p))/(C_(<a href="https://interviewquestions.tuteehub.com/tag/v-722631" style="font-weight:bold;" target="_blank" title="Click to know more about V">V</a>))` <br/> `:gt gamma -1 = (C_(p))/(C_(v))-1= ( C_(p)-C_(v))/(C_(v))= (R)/(C_(v))` ,brgt <a href="https://interviewquestions.tuteehub.com/tag/hence-484344" style="font-weight:bold;" target="_blank" title="Click to know more about HENCE">HENCE</a> `((V_(1))/(V_(2)))^(R//C_(v))= (T_(2))/(T_(1))` <br/> or `ln.(T_(2))/(T_(1))=(R)/(C_(v))ln.(V_(1))/(V_(2)) ` or `log.(T_(2))/(T_(1))= (R)/(C_(v)) log. (V_(1))/(V_(2))` <br/> i.e.,`log. (T_(2))/(300) = ( 8.314)/( 12.48) log. ( 12.5)/(25)` <br/> or ` logT_(2) = log 300 + 0.666 log . ( 1)/(2)` <br/> `= 2.4771 + 0.666 ( - 0.3010) = 2.2766` <br/> or`T_(2) =` Antilog 2.2766 `=<a href="https://interviewquestions.tuteehub.com/tag/189-1799650" style="font-weight:bold;" target="_blank" title="Click to know more about 189">189</a> K`<br/> `C_(p)=(DeltaH)/( DeltaT)`<br/>or `DeltaH =C_(p) Delta T ` or for <a href="https://interviewquestions.tuteehub.com/tag/n-568463" style="font-weight:bold;" target="_blank" title="Click to know more about N">N</a> moles `Delta H = n C_(p) Delta T`.<br/> Taking the gas as ideal. <br/> `PV = nRT ` or ` 1 xx 1.25 = n xx 0.0821 xx 300` or `n = 0.05` mole <br/> Further, `C_(p) = C_(v) + R = 12.48 + 8.314 = 20.794J` <br/> `:.Delta H = 0.05 xx 20.794 xx ( 189 - 300) = - 115 . 4 J`</body></html>
Discussion
49550.

A sample of argon gas at 1 atm pressure and 27^(@) C expand reversibly and adibatically from 1.25 dm^(3) to 2.50 dm^(3). The enthalpy change in this process will be C_(v.m.) for argon is 12.48 JK^(-1)mol^(-1).

Answer» <html><body><p>114.52 <a href="https://interviewquestions.tuteehub.com/tag/j-520843" style="font-weight:bold;" target="_blank" title="Click to know more about J">J</a><br/>`-117.14 J`<br/>`-57.26 J`<br/>57.26 J</p>Answer :D</body></html>
Discussion