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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.
| 50901. |
8 grams of NaOH are totally consumed to react with carbondioxide produced by the calcination of sodium bicarbonate. What weight of sodium bicarbonate is to be calcinated? |
| 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> :06.8g</body></html> | |
| 50902. |
8 gm O_(2) has the same number of molecules as in |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/11-267621" style="font-weight:bold;" target="_blank" title="Click to know more about 11">11</a> g `CO_(2)`<br/>22g `CO_(2)`<br/>7g CO<br/>14g CO</p>Solution :`8gm O_(2)=(8)/(32)=(1)/(4)` <a href="https://interviewquestions.tuteehub.com/tag/moles-1100459" style="font-weight:bold;" target="_blank" title="Click to know more about MOLES">MOLES</a> <br/> same molecules <a href="https://interviewquestions.tuteehub.com/tag/implies-1037962" style="font-weight:bold;" target="_blank" title="Click to know more about IMPLIES">IMPLIES</a> same moles <br/> `(11)/(<a href="https://interviewquestions.tuteehub.com/tag/44-316683" style="font-weight:bold;" target="_blank" title="Click to know more about 44">44</a>)=(1)/(4),(7)/(28)=(1)/(4)`</body></html> | |
| 50903. |
8 gm NaOH dissolve in 250 mL solution. This solution is diluted by using 500 ml water. Find the molartiy in' dilute solution. Also find moles of NaOH ? |
| Answer» <html><body><p> <br/> <br/> <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> :0.4 M and 0.2 <a href="https://interviewquestions.tuteehub.com/tag/mole-1100299" style="font-weight:bold;" target="_blank" title="Click to know more about MOLE">MOLE</a></body></html> | |
| 50904. |
8 gm H_2 and 256 gm HI in 4 L flask calculate this active mass. (H = 1 g "mol"^(-1), I = 127 g "mol"^(-1)) |
| 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/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a> <a href="https://interviewquestions.tuteehub.com/tag/mol-1100196" style="font-weight:bold;" target="_blank" title="Click to know more about MOL">MOL</a> `<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) H_2`, 0.5 mol `L^(-1)` <a href="https://interviewquestions.tuteehub.com/tag/hi-479908" style="font-weight:bold;" target="_blank" title="Click to know more about HI">HI</a></body></html> | |
| 50905. |
8 g of sulphur is burnt to form SO_(2) which is oxidised by chlorine water . The solution is treated with BaCl_(2) solution . The amount of BaSO_(4) precipitated is |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a> <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/>0.5 mole <br/>0.25 mole <br/>0.125 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/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a></body></html> | |
| 50906. |
7g of nitrogen occupies a volume of 5 litres under certain conditions. Under the same conditions one mole of a gas, having molecular weight 56, occupies a volume of |
| Answer» <html><body><p>40L<br/>20L<br/>10L<br/>80L</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> | |
| 50907. |
7g of nitrogen contains the same number of molecules as |
| Answer» <html><body><p>8g of oxygen <br/><a href="https://interviewquestions.tuteehub.com/tag/16g-277851" style="font-weight:bold;" target="_blank" title="Click to know more about 16G">16G</a> of oxygen <br/>8g of <a href="https://interviewquestions.tuteehub.com/tag/carbon-16249" style="font-weight:bold;" target="_blank" title="Click to know more about CARBON">CARBON</a> monoxide <br/><a href="https://interviewquestions.tuteehub.com/tag/22g-1825192" style="font-weight:bold;" target="_blank" title="Click to know more about 22G">22G</a> of carbon dioxide </p>Answer :A</body></html> | |
| 50908. |
7g of N_2 is allowed to react with 2g H_2 in presence of catalyst to form NH_3 Which is the limiting reagent? |
| 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> :`N_2`-<a href="https://interviewquestions.tuteehub.com/tag/limiting-2791961" style="font-weight:bold;" target="_blank" title="Click to know more about LIMITING">LIMITING</a> <a href="https://interviewquestions.tuteehub.com/tag/reagent-1178480" style="font-weight:bold;" target="_blank" title="Click to know more about REAGENT">REAGENT</a>,</body></html> | |
| 50909. |
7g of N_2 is allowed to react with 2g H_2 in presence of catalyst to form NH_3 Calculate the maximum amount of NH_3 that can be formed in the reaction. |
| 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> :`NH_3=8.5g`</body></html> | |
| 50910. |
750 mL of nitrogen are collected over water at 25^@C and 740 mm pressure. If the aqueous tension at this temperature is 23.8 mm Hg, calculate the mass of the dry gas. |
| 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> :According to Dalton.s law of partial pressure, <br/> `P_(gas) = P_(obs). - " aqueous <a href="https://interviewquestions.tuteehub.com/tag/tension-1241727" style="font-weight:bold;" target="_blank" title="Click to know more about TENSION">TENSION</a> " = 740 - 23.8 = 716.2 mm Hg` <br/>The given gas thus occupies a volume of <a href="https://interviewquestions.tuteehub.com/tag/750-335306" style="font-weight:bold;" target="_blank" title="Click to know more about 750">750</a> mL at `25^@C " and " 716.2 mm Hg`. <br/> According to the ideal gas equation, <br/> `PV =nRT= m/M RT` <br/> `m=(PVM)/(RT)` <br/> In the present case, <br/> `P =716.2 mm = (716.2)/<a href="https://interviewquestions.tuteehub.com/tag/760-335611" style="font-weight:bold;" target="_blank" title="Click to know more about 760">760</a> =0.9424` atm <br/> `V=750 mL = 0.750 L, "" T=25+273 =298 K` <br/> `M = 28 " and " R =0.0821 L " atm " K^(-1) mol^(-1)` <br/> `m=(0.9424xx0.750xx28)/(0.0821xx298) = 0.809 g` <br/> Hence, the mass of the given gas = 0.809 g</body></html> | |
| 50911. |
7.5 g of an acid are dissolved per litre of the solution. 20 mL of this acid solution required 25 mL of N//15 NaOH solution for complete neutralisation. Calculate the equivalent mass of the acid. |
| Answer» <html><body><p></p>Solution :Calculation of the normality of the acid <br/> Volume of acid taken = 20 <a href="https://interviewquestions.tuteehub.com/tag/ml-548251" style="font-weight:bold;" target="_blank" title="Click to know more about ML">ML</a> <br/> volume of base required = <a href="https://interviewquestions.tuteehub.com/tag/25-296426" style="font-weight:bold;" target="_blank" title="Click to know more about 25">25</a> mL <br/> Normality of base `= N//15` <br/> Applying normality equation : `ubrace(N_(1)xxV_(1))_("Acid") equiv ubrace(N_(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> V_(2))_("Base")`<br/> `N_(1) xx (20 mL) = ((1)/(15) N) xx 25 mL` <br/> `N_(1) = ((1)/(15) N) xx ((25 mL))/((20 mL)) = (1)/(12) N` <br/> Step <a href="https://interviewquestions.tuteehub.com/tag/ii-1036832" style="font-weight:bold;" target="_blank" title="Click to know more about II">II</a>. Calculation of the equivalent mass of the acid <br/> Amount of acid dissolved per litre of solution = 7.5 g <br/> `:.` Strength of the acid `= 7.5 g L^(-1)` <br/> Equivalent mass of the acid `= ("Strength of the acid")/("Normality of acid") = ((7.5 g L^(-1)))/((1//12 N)) = 90.0`</body></html> | |
| 50912. |
7.5 g of a gas occupies a volume of 5.6 litres at 0^(@)C and 1 atm pressure. The gas is |
| Answer» <html><body><p>NO<br/>`N_(2)O`<br/>`<a href="https://interviewquestions.tuteehub.com/tag/co-920124" style="font-weight:bold;" target="_blank" title="Click to know more about CO">CO</a>`<br/>`CO_(2)`</p>Solution :7.5 g of gas occupies a volume of 5.6 liters at 273 <a href="https://interviewquestions.tuteehub.com/tag/k-527196" style="font-weight:bold;" target="_blank" title="Click to know more about K">K</a> and 1 atm pressure. <a href="https://interviewquestions.tuteehub.com/tag/therefore-706901" style="font-weight:bold;" target="_blank" title="Click to know more about THEREFORE">THEREFORE</a>, the mass of gas that occupies a volume of 22.4 litres <br/>`(7.5g)/(5.6L) xx22.4L=<a href="https://interviewquestions.tuteehub.com/tag/30g-1853286" style="font-weight:bold;" target="_blank" title="Click to know more about 30G">30G</a>` <br/>Molar mass of NO (14+16) = 30g</body></html> | |
| 50913. |
7.5 g of a gas occupies 5.6 litres as STP. The gas is |
| Answer» <html><body><p>`NO`<br/>`N_(<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`<br/>`<a href="https://interviewquestions.tuteehub.com/tag/co-920124" style="font-weight:bold;" target="_blank" title="Click to know more about CO">CO</a>`<br/>`CO_(2)`</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> | |
| 50914. |
74.5 g metal chloride contain 35.5 g chlorine. Then the equlvalent weight of metal is..... |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/39-310302" style="font-weight:bold;" target="_blank" title="Click to know more about 39">39</a><br/>74.5<br/>78<br/>19.5</p>Solution :`74.5 - 35.5 = 39 <a href="https://interviewquestions.tuteehub.com/tag/g-1003017" style="font-weight:bold;" target="_blank" title="Click to know more about G">G</a>`</body></html> | |
| 50915. |
7.38 g of a sample of a metal oxide is quantitatively reduced to 6.84 g of pure metal. If the specific heat of the metal is 0.0332 cal/g, calculate the valency and the accurate atomic weight of the metal. |
| Answer» <html><body><p></p>Solution :Let the formula of the oxide be `M_(2)O_(x)` , where x is the valency of the <a href="https://interviewquestions.tuteehub.com/tag/metal-1094457" style="font-weight:bold;" target="_blank" title="Click to know more about METAL">METAL</a> M. We have, therefore, <br/> `x xx `moles of M = `2 xx` moles of O.<br/>We know that, atomic weight `xx` specific <a href="https://interviewquestions.tuteehub.com/tag/heat-21102" style="font-weight:bold;" target="_blank" title="Click to know more about HEAT">HEAT</a> `~~` 6.4 <br/> (Dulong and Petit.s law) <br/> `:.` approximate atomic weight `= (6.4)/(0.0332) = 193 ` <br/> From <a href="https://interviewquestions.tuteehub.com/tag/eqn-973463" style="font-weight:bold;" target="_blank" title="Click to know more about EQN">EQN</a>. `(<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>), x xx (6.64)/(193) =2xx ((7.38-6.84))/(16)`<br/> or <br/>x= 1.9. <br/> But valency is always a whole number and so the valency of the metal is 2. Now, to calculate the accurate value of the atomic weight of the metal, substitute the value of x as 2 in equation (1) again, <br/> or `2xx(6.84)/("atomic <a href="https://interviewquestions.tuteehub.com/tag/wt-1462289" style="font-weight:bold;" target="_blank" title="Click to know more about WT">WT</a>.") = 2xx((7.38 - 6.84))/(16)`<br/>Atomic weight (accurate) = 202.67.</body></html> | |
| 50916. |
723 KJ mol""^(-1) heat is released when 1 molemethanol combustions in present of O_(2). What amount of heat is enlisted of one mole of O_(2) is used? |
| Answer» <html><body><p>241 kJ<br/>723 kJ<br/>482 kJ<br/>924 KJ</p>Solution :Combusion reaction of methanol, <br/> `CH_(3) OH_((l)) + (3)/(2) O_(2(g)) = CO_(2(g)) + 2H_(2) O_((l)) , <a href="https://interviewquestions.tuteehub.com/tag/delta-947703" style="font-weight:bold;" target="_blank" title="Click to know more about DELTA">DELTA</a> <a href="https://interviewquestions.tuteehub.com/tag/h-1014193" style="font-weight:bold;" target="_blank" title="Click to know more about H">H</a>= -723` kJ <br/> 1 mole `(3)/(2)` mole <br/> therefore `(3)/(2)` moles `O_2` is used in this reaction. <br/> If 1 mole `O_2` is used. <br/> then free <a href="https://interviewquestions.tuteehub.com/tag/energy-15288" style="font-weight:bold;" target="_blank" title="Click to know more about ENERGY">ENERGY</a> `= (2)/(3) xx` combustion energy 723 <br/> `= 482` kJ</body></html> | |
| 50917. |
720cc of methane diffused through a porous membrane in 30min. Under identical conditions 240cc of gas 'X'diffused in 20 min. Calculate the molecular weight of 'X'. |
| Answer» <html><body><p></p>Solution :Rate of diffusion of methane ` = 720//30 = 24 "cc" m^(-<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/> Rate of diffusion of gas .<a href="https://interviewquestions.tuteehub.com/tag/x-746616" style="font-weight:bold;" target="_blank" title="Click to know more about X">X</a>.`240//20 = 12 "cc" m^(-1)` <br/> Ratio of rates of diffusion is <a href="https://interviewquestions.tuteehub.com/tag/given-473447" style="font-weight:bold;" target="_blank" title="Click to know more about GIVEN">GIVEN</a> as `(r_(1))/(r_(2))= sqrt((M_(2))/(M_(1))) = (24)/(12) = 12` <br/> Molecular weight of .X. ` = 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> 16 = 64`</body></html> | |
| 50918. |
7.18 g iron displaces 2.04 g copper from copper sulphate solution. If the equivalent weight of copper is 31.7, calculate the equivalent weight of iron. |
| Answer» <html><body><p></p>Solution :In the present case, <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 <a href="https://interviewquestions.tuteehub.com/tag/iron-1051344" style="font-weight:bold;" target="_blank" title="Click to know more about IRON">IRON</a> taken = 1.8 g <br/>Mass of copper displaced = 2.04 g <br/> Equivalent mass of copper = 31.7<br/> Using the relation `=("Mass of iron")/("Mass of copper") = ("Equivalent weight of iron")/("Equivalent weight of copper") ` <br/> and substituting the <a href="https://interviewquestions.tuteehub.com/tag/values-25920" style="font-weight:bold;" target="_blank" title="Click to know more about VALUES">VALUES</a>, we have <br/> `1.8/2.04 = ("Equivalent weight of iron")/31.7` <br/> or Equivalent weight of iron `=(1.8 <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> 31.7)/2.04= 27.97`</body></html> | |
| 50919. |
70 g of a sample of magnesite on treatment with excess of HCl gave 11.2 L of CO_(2) at STP. The percentage purify of the sample |
| Answer» <html><body><p>80<br/>70<br/>60<br/>50</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> | |
| 50920. |
7.00 g of a gas occupies a volume of 4.1 L at 300 K and 1 atm pressure. What is the molecular mass of the gas? |
| Answer» <html><body><p></p>Solution :In the <a href="https://interviewquestions.tuteehub.com/tag/present-1163722" style="font-weight:bold;" target="_blank" title="Click to know more about PRESENT">PRESENT</a> case, <br/> `P= 1 " atm, " V= 4.1 L, T = 300 <a href="https://interviewquestions.tuteehub.com/tag/k-527196" style="font-weight:bold;" target="_blank" title="Click to know more about K">K</a>, m= 7.0 g, " and " R=0.0821 L " atm " K^(-1) mol^(-1)` <br/>`:."" <a href="https://interviewquestions.tuteehub.com/tag/pv-593601" style="font-weight:bold;" target="_blank" title="Click to know more about PV">PV</a> = m/M RT` <br/>`:."" M=(mRT)/(PV)= (7.0 xx0.0821 xx 300)/(1 x 4.1)= 42` <a href="https://interviewquestions.tuteehub.com/tag/amu-25369" style="font-weight:bold;" target="_blank" title="Click to know more about AMU">AMU</a><br/> and `:. ""d = m/V` <br/> `:. "" d=(7.0)/(4.1)=1.7g L^(-1)` <br/> Hence, the molecular mass of the given gas is 42 amu and its density in the given conditions is`1.7 g L^(-1)`.</body></html> | |
| 50921. |
70 gms of a metal oxide on reduction produced 54 gms of metal. The atomic weight of the metal is 81. What is its valency? |
| Answer» <html><body><p><br/></p>Solution :wt. of metal oxide = 70 <a href="https://interviewquestions.tuteehub.com/tag/g-1003017" style="font-weight:bold;" target="_blank" title="Click to know more about G">G</a> <br/> wt. of metal = 54 g <br/> `therefore` wt of oxygen = `70-54=16g` <br/> `(W_(<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>))/(W_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>))=(E_(1))/(E_(2)),("wt of metal")/("wt of oxygen")=("EQ wt of metal")/("EQ wt of oxygen")` <br/> `(54)/(16)=(E)/(8)impliesE=27` <br/> At. Wt = EQ wt `xx` valency <br/> `81=27xx` valency implies Valency = 3</body></html> | |
| 50922. |
7 grams of nitrogen is present at 127^@C and 16 grams of oxygen at 27^@C. Calculate (a) ratio of kinetic energy and (b) ratio of average kinetic energy of nitrogen and oxygen. |
| Answer» <html><body><p></p>Solution :`(K.E) = 3/2 <a href="https://interviewquestions.tuteehub.com/tag/nrt-1109890" style="font-weight:bold;" target="_blank" title="Click to know more about NRT">NRT</a> ` <br/>For two gases ratio of <a href="https://interviewquestions.tuteehub.com/tag/kinetic-533291" style="font-weight:bold;" target="_blank" title="Click to know more about KINETIC">KINETIC</a> energies = <br/> ` ((K.E )_1)/((K.E)_2) = (n_1 T_1)/(n_2 T_2)= (<a href="https://interviewquestions.tuteehub.com/tag/7-332378" style="font-weight:bold;" target="_blank" title="Click to know more about 7">7</a>)/(28 ) xx 400 xx(32)/(16) xx (1)/(300 ) = 2:3`</body></html> | |
| 50923. |
7 g of a sample of sodium chloride on treatment with excess of silver nitrate gave 14.35 g of AgCl. The percentage of NaCl in the sample is |
| Answer» <html><body><p>80<br/>50<br/>65.8<br/>83.5</p>Answer :D</body></html> | |
| 50924. |
7-Bromo-1,3,5-cycloheptatriene exists as an ion whereas 5-bromo-1,3-cyclopentadiene does not form an ion even in presence of Ag^+. Explain |
| Answer» <html><body><p></p>Solution :7-Bromo-1,3,5-cycloheptatriene, on ionization, gives tropylium ion. Since , tropylium ion contains `6 <a href="https://interviewquestions.tuteehub.com/tag/pi-600185" style="font-weight:bold;" target="_blank" title="Click to know more about PI">PI</a>`-electrons which are completely delocalized, therefore, according to Huckel rule, it is <a href="https://interviewquestions.tuteehub.com/tag/aromatic-363924" style="font-weight:bold;" target="_blank" title="Click to know more about AROMATIC">AROMATIC</a> and <a href="https://interviewquestions.tuteehub.com/tag/hence-484344" style="font-weight:bold;" target="_blank" title="Click to know more about HENCE">HENCE</a> stable .Being highly stable , it is easily <a href="https://interviewquestions.tuteehub.com/tag/formed-464209" style="font-weight:bold;" target="_blank" title="Click to know more about FORMED">FORMED</a> . <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/PR_CHE_02_XI_C13_E19_002_S01.png" width="80%"/> <br/> In contrast,5-bromo-1,3-cyclopentadiene, on ionization, will give 1,3-cyclopentadienyl cation which contains `4pi`-electrons and hence in antiaromatic. Being antiaromatic, it is highly unstable and hence is not formed even in the presence of `Ag^+` ion which otherwise facilitates ionization.</body></html> | |
| 50925. |
6xx10^(22) gas molecules each of mass 10^(-24)kg are taken in a vessel of 10 litre. What is the pressure exerted by gas molecules? The average velocity of the gas molecules is 92.62m/sec. |
| Answer» <html><body><p>`<a href="https://interviewquestions.tuteehub.com/tag/2xx10-1840192" style="font-weight:bold;" target="_blank" title="Click to know more about 2XX10">2XX10</a>^(<a href="https://interviewquestions.tuteehub.com/tag/5-319454" style="font-weight:bold;" target="_blank" title="Click to know more about 5">5</a>)<a href="https://interviewquestions.tuteehub.com/tag/pa-1145246" style="font-weight:bold;" target="_blank" title="Click to know more about PA">PA</a>`<br/>20Pa<br/>`2xx10^(<a href="https://interviewquestions.tuteehub.com/tag/6-327005" style="font-weight:bold;" target="_blank" title="Click to know more about 6">6</a>)Pa`<br/>`2xx10^(4)Pa`</p>Answer :d</body></html> | |
| 50926. |
6L H_2O_2 contains 440 gm solute than how much litre O_2 gas will produce from this solution at STP ? |
| Answer» <html><body><p>24.16<br/>22.8<br/>30.16<br/>25.8</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> :Molarity=`"Weight of solute <a href="https://interviewquestions.tuteehub.com/tag/gm-1008640" style="font-weight:bold;" target="_blank" title="Click to know more about GM">GM</a>"/"Molecular massx Volume of solution in litre"` <br/> `=<a href="https://interviewquestions.tuteehub.com/tag/440-316801" style="font-weight:bold;" target="_blank" title="Click to know more about 440">440</a>/(34xx6)=2.156 M H_2O_2` <br/> `V(O)_2`= M x 11.2 <br/> = 2.156 x 11.2 <br/> =24.16 L `O_2` <a href="https://interviewquestions.tuteehub.com/tag/gas-1003521" style="font-weight:bold;" target="_blank" title="Click to know more about GAS">GAS</a></body></html> | |
| 50927. |
6g. of Urea is dissolved in 90g. of water. The mole fraction of solute is |
| Answer» <html><body><p>`1//5`<br/>`1//50`<br/>`1//51`<br/>`1//501`</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/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a></body></html> | |
| 50928. |
6g of NaOH are dissolved in 200 cm^3 of water. What is the relation between molarity and normality of the solution thus obtained ? |
| 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/molarity-1100268" style="font-weight:bold;" target="_blank" title="Click to know more about MOLARITY">MOLARITY</a> and <a href="https://interviewquestions.tuteehub.com/tag/normality-1124423" style="font-weight:bold;" target="_blank" title="Click to know more about NORMALITY">NORMALITY</a> are the same</body></html> | |
| 50929. |
6g of Mg reacts with excess of an acid. The amount of hydrogen produced would be |
| Answer» <html><body><p>0.5g<br/>1g<br/>2g<br/>4g</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> | |
| 50930. |
6g of hydrogen at 1.5 atm and 273^@C occupies a volume of |
| Answer» <html><body><p>44.8L <br/>89.6L <br/>67.2L <br/>11.2L </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> | |
| 50931. |
""_(6)C^(12) and ""_(1)T^(3) are formed in nature due to the nuclear reaction of neutron with |
| Answer» <html><body><p>`""_(7)N^(14)`<br/>`""_(6)<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a>^(13)`<br/>`""_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)He^(<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/>`""_(3)Li(6)`</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> :`""_(7)^(14)N+_(0)^(1)n to ""_(6)^(12)C+""_(1)^(3)T`</body></html> | |
| 50932. |
6.9g N_2O_4 is taken 0.5 L closed vessel at 400 K temperature. The equilibrium N_2O_(4(g)) hArr 2NO_(2(g)) total pressure at equilibrium is 9.15 atm calculate K_c, K_p and partial pressure of each component. |
| 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> :`p_(N_2O_4)`= 0.69 atm , `p_(NO_2)`= 8.46 atm , `K_p`=103.73 atm , `K_c=3.162 "<a href="https://interviewquestions.tuteehub.com/tag/mol-1100196" style="font-weight:bold;" target="_blank" title="Click to know more about MOL">MOL</a> <a href="https://interviewquestions.tuteehub.com/tag/l-535906" style="font-weight:bold;" target="_blank" title="Click to know more about L">L</a>"^(-<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>)`</body></html> | |
| 50933. |
6.8 gm of H_2O_2 is present in 500 ml of an aqueous solution. When 10 ml of this solution is decomposed, what is the S.T.P. volume of oxygen obtained. |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/560-325686" style="font-weight:bold;" target="_blank" title="Click to know more about 560">560</a> <a href="https://interviewquestions.tuteehub.com/tag/ml-548251" style="font-weight:bold;" target="_blank" title="Click to know more about ML">ML</a><br/>11.2 ml<br/>44.8 ml<br/>22.4 ml </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/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a></body></html> | |
| 50934. |
66 gm sample of an oxalate salt Al_(x)H_(y)(C_(2)O_(4))_(z).nH_(2)O is dissolved in water to form 500mL solution. 50mL solution requires 60mL of 0.5M Ba(OH)_(2) and 240mL 0.1 M KMnO_(4) in acidic medium separately. If in salt x,y,z and n are present in simplest ratio, then select the correct statement(s). |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/moles-1100459" style="font-weight:bold;" target="_blank" title="Click to know more about MOLES">MOLES</a> of oxalate salt in original sample is 0.2<br/>Ratio of y/z is <a href="https://interviewquestions.tuteehub.com/tag/equal-446400" style="font-weight:bold;" target="_blank" title="Click to know more about EQUAL">EQUAL</a> is 1<br/>Value of (x+y+z)-n=5<br/>Number of oxalate <a href="https://interviewquestions.tuteehub.com/tag/ion-1051153" style="font-weight:bold;" target="_blank" title="Click to know more about ION">ION</a> per molecule of oxalate salt is 2</p>Answer :a,b,c</body></html> | |
| 50935. |
6.54 g of zinc are treated with 11.5 g of H_2SO_4. Calculate the volume of hydrogen evolved at S.T.P. How much H_2SO_4 will be left in excess? |
| Answer» <html><body><p></p>Solution :The <a href="https://interviewquestions.tuteehub.com/tag/corresponding-935567" style="font-weight:bold;" target="_blank" title="Click to know more about CORRESPONDING">CORRESPONDING</a> <a href="https://interviewquestions.tuteehub.com/tag/chemical-914796" style="font-weight:bold;" target="_blank" title="Click to know more about CHEMICAL">CHEMICAL</a> equation is: <br/> `underset(65.38 g)(Zn) + underset(2.016 + 32.06 + 64.0= 98.1 g)(H_(2)SO_(4)) to ZnSO_(4) + underset(22.4 "L at S.T.P")(H_(2))` <br/> From the equation, it is clear that 6.54 g of zinc will react with 9.81 g of `H_2SO_4`. Thus, the amount of `H_2SO_4` <a href="https://interviewquestions.tuteehub.com/tag/taken-659096" style="font-weight:bold;" target="_blank" title="Click to know more about TAKEN">TAKEN</a> is in excess. Hence, Zn will get consumed completely and this is the limiting reagent. <br/> `therefore 65.38 g` of zinc <a href="https://interviewquestions.tuteehub.com/tag/evolve-977729" style="font-weight:bold;" target="_blank" title="Click to know more about EVOLVE">EVOLVE</a> `H_(2) = 22.4 L` at S.T.P. <br/> `therefore 6.54 g` of zinc will evolve <br/> `H_(2) = 22.4/(65.38) xx 6.54 = 2.24` L at S.T.P. <br/> Further, <br/> `therefore 65.38` g of zinc react with `=H_(2)SO_(4) = 98.1 g` <br/> `therefore H_(2)SO_(4)` left in excess `=11.5- 98.1 = 1.69 g` <br/> Hence, 2.24 L hydrogen will be evolved at S.T.R and 1.69 g of `H_2SO_4` will be left in excess.</body></html> | |
| 50936. |
.^(64)Cu (hlaf lifre = 12.8h)deacays by beta^(-) emission (38%), beta^(+)- emission (19%) and electron capute (43%). Write the decay products and calculatepartial half-live for each of the decay processes. |
| Answer» <html><body><p></p>Solution :We know,<br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/DPB_PHY_CHM_IX_C04_E01_051_S01.png" width="80%"/> <br/> Given `lambda_(av) = (0.693)/(12.8) hr^(-1)`<br/>`:. lambda_(1) +lambda_(2) +lambda_(3) =lambda_(av) = (0.693)/(12.8) = 5.41xx10^(-2) hr^(-1)` .....(1) <br/> <a href="https://interviewquestions.tuteehub.com/tag/also-373387" style="font-weight:bold;" target="_blank" title="Click to know more about ALSO">ALSO</a> for parallel <a href="https://interviewquestions.tuteehub.com/tag/path-11833" style="font-weight:bold;" target="_blank" title="Click to know more about PATH">PATH</a> <a href="https://interviewquestions.tuteehub.com/tag/decay-945588" style="font-weight:bold;" target="_blank" title="Click to know more about DECAY">DECAY</a> <br/> `lambda_(1)` = Fractionlal yield of `._(30)^(64)Zn xx lambda_(av)` <br/> `lambda_(2)` = Fractionalyeild of `._(28)^(64) Ni xx lambda_(av)`<br/> `lambda_(3)` = Fractionlal yield of `._(28)^(64) Ni xx lambda_(av)` <br/> `:. (lambda_(1))/(lambda_(2)) = (38)/(19)` ....(2) <br/> `(lambda_(1))/(lambda_(3)) = (38)/(43)` ....(3) <br/> From eqs. (1), (2) and (3) <br/> `lambda_(1) = 2.056xx10^(-2) hr^(-1)` <br/> `lambda_(2) = 1.028xx10^(-2) hr^(-1)` <br/> `lambda_(3) = 2.327xx10^(-2) hr^(-1)` <br/> `:. t_(1//2)` for `beta^(-)` emission <br/> `= (0.693)/(2.056xx10^(-2)) = 33.70 hr` <br/>`t_(1//2)` for `beta^(+)` emission <br/> `= (0.693)/(1.028xx10^(-2)) = 67.41 hr` <br/>`t_(1//2)` for electron capture <br/> `= (0.693)/(2.327xx10^(-2)) = 29.28 hr`</body></html> | |
| 50937. |
6.46gm BaSO_(4) is obtain from 4.81gm compound in estimation of sulphur by Carius method. Calculat ethe % of sulphur |
| 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> :0.185</body></html> | |
| 50938. |
64 gm of CH_(4) "and" 68gm "of" H_(2)S was placed in an close container and heated up to 727^(@)C following equilibrium is established in gaseous phase reaction is: CH_(4)(g)+2H_(2)S(g)hArrCS_(2)(g)+4H_(2)(g) The total pressure at equilibrium is 1.6 atm and partial pressure of H_(2) "is" 0.8 atm. Then |
| Answer» <html><body><p>Total moles at equilibrium `4.8`<br/>`K_(P)=K_(C)(<a href="https://interviewquestions.tuteehub.com/tag/rt-615359" style="font-weight:bold;" target="_blank" title="Click to know more about RT">RT</a>)^(2)`<br/>Mole fraction `H_(2)` at equilibrium`=0.5`<br/>On increasing moles of `H_(2)S` equilibrium constant increases.</p>Solution :`CH_(4)(g)+2H_(2)S(g)hArrCS_(2)(g)+4H_(2)(g)` <br/> `{:(t=0,64gm,68gm),("Moles",4"mole",2"mole"):}` <br/> `4-x 2-2x x 4x` <br/> Total moles at equilibrium`=4-x+2-2x+x+4x=(6-2x)` <br/> <a href="https://interviewquestions.tuteehub.com/tag/acc-846605" style="font-weight:bold;" target="_blank" title="Click to know more about ACC">ACC</a>. To `PV=nRT,T=727+273=1000K` <br/> `1.84xxV=(6-2x)xxRxx1000 `....(1) <br/> For `H_(2) "gas" V,P_(H_(2))=n_(H_(2))RTto0.8xxV=4xxRxx1000` ....(2) <br/> From <a href="https://interviewquestions.tuteehub.com/tag/eq-446394" style="font-weight:bold;" target="_blank" title="Click to know more about EQ">EQ</a> (1) and(2) `(1.6)/(0.8)=((6-2x))/(4x)` <br/> `x=0.6`</body></html> | |
| 50939. |
6.3 g of oxalic acid dihydrate have been dissolved in water to obtain a 250 ml solution. How much volume of 0.1 N NaOH would be required to neutralize 10 mL of this solution ? |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/40-314386" style="font-weight:bold;" target="_blank" title="Click to know more about 40">40</a> mL<br/>20 mL<br/>10 mL<br/>4 mL </p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/eq-446394" style="font-weight:bold;" target="_blank" title="Click to know more about EQ">EQ</a>. wt of `overset(COOH)underset(COOH)(|).2H_(2)O = 63`<br/> Normality of the solution prepared <br/> `N=(w xx 1000)/(E xx <a href="https://interviewquestions.tuteehub.com/tag/v-722631" style="font-weight:bold;" target="_blank" title="Click to know more about V">V</a>) = (6.3 xx 1000)/(63 xx 250) = 0.4 N` <br/> `underset(NaOH)(N_(1)V_(1)) = underset("<a href="https://interviewquestions.tuteehub.com/tag/oxalic-2906163" style="font-weight:bold;" target="_blank" title="Click to know more about OXALIC">OXALIC</a> acid")(N_(2)V_(2))` <br/> `therefore V_(1) = (N_(2)V_(2))/N_(1) =(0.4 xx 10)/0.1 = 40 mL`</body></html> | |
| 50940. |
60mL of pure ozone is heated and then colled. The increase in the volume is 10mL. Determine the percentage volume of ozone decomposed into oxygen. |
| Answer» <html><body><p></p>Solution :The balanced equations for the decomposed of ozone is <br/> `2O_(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)(g)to3O_(<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)` <br/> `2 vol "" 3vol"Gay-Lussac.s <a href="https://interviewquestions.tuteehub.com/tag/coefficients-920961" style="font-weight:bold;" target="_blank" title="Click to know more about COEFFICIENTS">COEFFICIENTS</a>"` <br/> `60mL "" "0mL at start"` <br/> `60-2x "" 3x "after the <a href="https://interviewquestions.tuteehub.com/tag/reaction-22747" style="font-weight:bold;" target="_blank" title="Click to know more about REACTION">REACTION</a>"` <br/> Total volume of the mixture after the reaction =60-2x+3x=60+10 <br/> <a href="https://interviewquestions.tuteehub.com/tag/solving-1217196" style="font-weight:bold;" target="_blank" title="Click to know more about SOLVING">SOLVING</a>, x=10 <br/> Volume of `O_(3)` decomposed =20mL <br/> Percentage volume of `O_(3)` decomposed `=(20/(60)xx100=33.3%)`</body></html> | |
| 50941. |
60g CH_(3)COOH and 46g C_(2)H_(5)OH react in 5L flask to form 44g CH_(3)COOC_(2)H_(5) at equilibrium. On taking 120 g CH_(3)COOH and 46g C_(2)H_(5)OH, CH_(3)COOC_(2)H_(5) formed at equilibrium is : |
| Answer» <html><body><p>44g<br/>20.33g<br/>22g<br/>58.66g</p>Solution :Molar mass of `CH_(3)COOH=60gmol^(-1)` <br/> Molar mass of `C_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)H_(5)OH=46g mol^(-1)` <br/> Molar mass of `CH_(3)COOC_(2)H_(5)=88g mol^(-1)` <br/> `:. [CH_(3)COOH)_("Initial")=(<a href="https://interviewquestions.tuteehub.com/tag/60-328817" style="font-weight:bold;" target="_blank" title="Click to know more about 60">60</a>)/(60xx5)=0.2 "mol"L^(-1)` <br/> `[C_(2)H_(5)OH]_("Initial")=(46)/(46xx5)=0.2"mol" L^(-1)` <br/> `[CH_(3)COOC_(2)H_(5)]_(eqm)=(44)/(88)xx(1)/(5)=0.1 "mol" L^(-1)` <br/> `CH_(3)COOH+C_(2)H_(5)OH hArr CH_(3)COOC_(2)H_(5)+H_(2)O` <br/> Initial <br/> `0.2 M ""0.2M` <br/> At eqm. <br/> `(0.2-0.1)M""(0.2-0.1)M""0.1M""0.1M` <br/> `:. K=([CH_(3)COOC_(2)H_(5)][H_(2)O])/([CH_(3)COOH][C_(2)H_(5)OH])` <br/> `=(0.1xx0.1)/(0.1xx0.1)=1` <br/>In second case, <br/> `[CH_(3)COOH]_("Initial")=0.4M` <br/> `[C_(2)H_(5)OH]_("Initial")=0.2M` <br/><br/> If x is the amount of acid and <a href="https://interviewquestions.tuteehub.com/tag/alcohol-853024" style="font-weight:bold;" target="_blank" title="Click to know more about ALCOHOL">ALCOHOL</a> reacted <br/> `[CH_(3)COOH]_("eqm.")=(0.40-x)M` <br/> `[C_(2)H_(5)OH]_("eqm.")=[0.2-x]M` <br/> `[CH_(3)COOC_(2)H_(5)]_("eqm.")=[H_(2)O]_("eqm.")=xM` <br/> `:. K=(x^(2))/((0.4-x)(0.2-x))=1` <br/> `:. x^(2)=(0.4-x)(0.2-x)` <br/> `x^(2)=0.08+x^(2)-0.6x` <br/> `0.6x=0.08` <br/> `x=(8)/(60)M` <br/> `:. `Moles of ethyl acetate produced `=(8)/(60)<a href="https://interviewquestions.tuteehub.com/tag/xx5-2340058" style="font-weight:bold;" target="_blank" title="Click to know more about XX5">XX5</a>=(2)/(3)` <br/> Mass of ethyl acetate produced <br/> `=(2)/(3)xx88=58=58.66g`.</body></html> | |
| 50942. |
60g of a compound on analysis gave C=24g, H=4g and O=32g. Its empirical formula is |
| Answer» <html><body><p>`C_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)H_(<a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>)O_(2)`<br/>`C_(2)H_(2)O_(2)`<br/>`CH_(2)O_(2)`<br/>`CH_(2)O`</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> :D</body></html> | |
| 50943. |
6.048 g H_(2) and 28 g N_(2)react with each other and produce 34.048 g NH_(3). This reaction is explained by which law ? |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/constant-930172" style="font-weight:bold;" target="_blank" title="Click to know more about CONSTANT">CONSTANT</a> <a href="https://interviewquestions.tuteehub.com/tag/proportion-1170266" style="font-weight:bold;" target="_blank" title="Click to know more about PROPORTION">PROPORTION</a><br/> <a href="https://interviewquestions.tuteehub.com/tag/multiple-1105557" style="font-weight:bold;" target="_blank" title="Click to know more about MULTIPLE">MULTIPLE</a> proportion<br/> Combining weights<br/> Charles law</p>Answer :A::<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> | |
| 50944. |
6.023xx10^(22) atoms of Hydrogen can make |
| Answer» <html><body><p>0.05 <a href="https://interviewquestions.tuteehub.com/tag/moles-1100459" style="font-weight:bold;" target="_blank" title="Click to know more about MOLES">MOLES</a> of `H_(2)` molecules <br/>0.1 gms of <a href="https://interviewquestions.tuteehub.com/tag/hydrogens-1034293" style="font-weight:bold;" target="_blank" title="Click to know more about HYDROGENS">HYDROGENS</a> <a href="https://interviewquestions.tuteehub.com/tag/atoms-887421" style="font-weight:bold;" target="_blank" title="Click to know more about ATOMS">ATOMS</a> <br/>0.1 <a href="https://interviewquestions.tuteehub.com/tag/gram-1010695" style="font-weight:bold;" target="_blank" title="Click to know more about GRAM">GRAM</a>-molecules of Hydrogen <br/>0.1 gram atoms of Hydrogen</p>Solution :1 mole of compound `-6.023xx10^(23)` molecules <br/> Hydrogen `(H_(2))` is diatomic <br/> 1 gram atom of any <a href="https://interviewquestions.tuteehub.com/tag/element-969236" style="font-weight:bold;" target="_blank" title="Click to know more about ELEMENT">ELEMENT</a> <br/> `=6.023xx10^(23)` atoms of that element <br/> Atomic weight of hydrogen = 1 <br/> Number of gram atoms = `("Weight in grams")/("Gramotomicweight")`</body></html> | |
| 50945. |
6.023xx10^(20) molecules of urea are present in 100 ml of its solution. The concentration of the solution is |
| Answer» <html><body><p>0.001 M<br/>0.01 M<br/>0.02 M<br/>0.1 M</p>Solution :100 ml of solution contain `6.023 <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> 10^(20)` molecules. <br/>`6.023 xx 10^(23)` molecules in 1000 ml = 1 M <br/>`:. 6.023 xx 10^(20)` molecules in 1000 ml=`(1xx6.023xx10^(20))/(6.023xx10^(23))=10^(-3)` <br/>`10^(-3)` moles of <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 100 ml <br/>`:.` In 1000ml the moles present is=`10^(-3)/cancel(100)xxcancel(1000)=10^(-2)` moles <br/><a href="https://interviewquestions.tuteehub.com/tag/concentration-20558" style="font-weight:bold;" target="_blank" title="Click to know more about CONCENTRATION">CONCENTRATION</a> = 0.01 M</body></html> | |
| 50946. |
6.022 xx 10^(22) molecules of N_(2) at NTP will occupy a volume of |
| Answer» <html><body><p>22.4 <a href="https://interviewquestions.tuteehub.com/tag/litres-1075876" style="font-weight:bold;" target="_blank" title="Click to know more about LITRES">LITRES</a> <br/>2.24 litres <br/>6.02 litres<br/>6.02 <a href="https://interviewquestions.tuteehub.com/tag/ml-548251" style="font-weight:bold;" target="_blank" title="Click to know more about ML">ML</a> </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> | |
| 50947. |
6.022 xx 10^(20) molecules of urea are present in 100 mL of its solution. The concentration of urea solution is : |
| Answer» <html><body><p>0.001 M<br/>0.01 M<br/>0.02 M<br/>0.1 M</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. of <a href="https://interviewquestions.tuteehub.com/tag/moles-1100459" style="font-weight:bold;" target="_blank" title="Click to know more about MOLES">MOLES</a> urea <br/>`(6.022xx10^(20))/(6.022xx10^(<a href="https://interviewquestions.tuteehub.com/tag/23-294845" style="font-weight:bold;" target="_blank" title="Click to know more about 23">23</a>))=0.001` <a href="https://interviewquestions.tuteehub.com/tag/mol-1100196" style="font-weight:bold;" target="_blank" title="Click to know more about MOL">MOL</a> <br/> Concentration of solution <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>^(-3)"mol"))/(("0.1 L"))xx10^(-2)"molL"^(-1)=0.01 M`.</body></html> | |
| 50948. |
6.02 xx 10^(20) molecules of urea are present in 100 mL of its solution. The concentration of urea solution is: |
| Answer» <html><body><p>0.001 M<br/>0.01 M<br/>0.02 M<br/>0.1 M </p>Solution :No. of <a href="https://interviewquestions.tuteehub.com/tag/moles-1100459" style="font-weight:bold;" target="_blank" title="Click to know more about MOLES">MOLES</a> of urea taken `=(6.02 xx 10^(20))/(6.022 xx 10^(23)) = 10^(-<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)` <br/> `therefore` <a href="https://interviewquestions.tuteehub.com/tag/molarity-1100268" style="font-weight:bold;" target="_blank" title="Click to know more about MOLARITY">MOLARITY</a> of solution `=10^(-3)/<a href="https://interviewquestions.tuteehub.com/tag/100-263808" style="font-weight:bold;" target="_blank" title="Click to know more about 100">100</a> xx 1000 = 0.01 M`</body></html> | |
| 50949. |
60 mL of methane and 180mL of oxygen are mixed and fixed. The resulting gases are passed throughcaustic soda solution. Determine the composition of gases before and after passing through base solution. |
| Answer» <html><body><p></p>Solution :The storichiometric equation is `underset((g))(CH_(4))+underset((g))(2O_(2))tounderset((g))(CO_(2))+underset((l))(2H_(2))O` <br/> `1 vol 2 vol 1vol ovol """Gay-Lussac.s coefficients"` <br/> `60mL 180mL 0mL 0mL """ at start"` <br/> `0mL 60mL 60mL 0mL "" "after the reaction" ` <br/> Before passing through caustic soda solution unreacted oxygen =60mL <br/> Carbondioxide <a href="https://interviewquestions.tuteehub.com/tag/formed-464209" style="font-weight:bold;" target="_blank" title="Click to know more about FORMED">FORMED</a> =<a href="https://interviewquestions.tuteehub.com/tag/80ml-1927563" style="font-weight:bold;" target="_blank" title="Click to know more about 80ML">80ML</a> <br/> After passing through caustic soda solution(since `CO_(2)` is <a href="https://interviewquestions.tuteehub.com/tag/absorbed-846166" style="font-weight:bold;" target="_blank" title="Click to know more about ABSORBED">ABSORBED</a>) <br/> unreaced oxygen =60mL</body></html> | |
| 50950. |
60 gm of oleum (labelled as 118%) is mixed with 11.8 gm of water. What will be the composition of final mixture? |
| Answer» <html><body><p>Only `H_(2)SO_(4)`, having <a href="https://interviewquestions.tuteehub.com/tag/mass-1088425" style="font-weight:bold;" target="_blank" title="Click to know more about MASS">MASS</a> 71.8 gm<br/>118 gm of `H_(2)SO_(4)`<br/>70.8 gm `H_(2)SO_(4)` and <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 water <br/><a href="https://interviewquestions.tuteehub.com/tag/32-307188" style="font-weight:bold;" target="_blank" title="Click to know more about 32">32</a> gm `SO_(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)` and 39.8 gm `H_(2)SO_(4)`</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> | |