InterviewSolution
Saved Bookmarks
InterviewSolution
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.
| 51751. |
10 Its of an air sample with relative humidity 0.6 is compressed to 5 litres at same temperature. What is the partial pressure of the water vapour in the compressed air, if the pressure of saturated water vapour at that temperature is 2.4 K Pa |
| Answer» <html><body><p>1.2 KPa <br/>2.4 Kpa <br/>3.6 kPa<br/>1.8 KPa <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> :`R.H. = 0.6 = (P_(H_2 O))/(V.P) <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` <br/> `implies P_(H_2 O) = 0.6 xx 2.4 = 1.44 kPa` <br/> on making volume half, `P_(H_2 O)` should double, but R.H. `> 100% implies` saturation. <br/> `implies P_(H_2O) = 2.4 kPa` (<a href="https://interviewquestions.tuteehub.com/tag/limiting-2791961" style="font-weight:bold;" target="_blank" title="Click to know more about LIMITING">LIMITING</a> value) .</body></html> | |
| 51752. |
10 grams of steam at 100^(@)C is mixed with 50 gm of ice at 0^(@)C then final temperture is |
| 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> :The kinetic energy of <a href="https://interviewquestions.tuteehub.com/tag/molecules-563030" style="font-weight:bold;" target="_blank" title="Click to know more about MOLECULES">MOLECULES</a> in <a href="https://interviewquestions.tuteehub.com/tag/steam-1226863" style="font-weight:bold;" target="_blank" title="Click to know more about STEAM">STEAM</a> will be <a href="https://interviewquestions.tuteehub.com/tag/greater-476627" style="font-weight:bold;" target="_blank" title="Click to know more about GREATER">GREATER</a> than the kinetic energy of molecules in <a href="https://interviewquestions.tuteehub.com/tag/ice-1035302" style="font-weight:bold;" target="_blank" title="Click to know more about ICE">ICE</a>. Hence internal energy of steam is greater than that of ice.</body></html> | |
| 51753. |
10 grams of argon gas is compressed isothermally and reversibly at a temperature of 27^(@)C from 10L to 5L. Calculate the work done for this process in multiple of 10^(20 calories. |
| Answer» <html><body><p><br/></p>Solution :`W_("<a href="https://interviewquestions.tuteehub.com/tag/isothermal-520014" style="font-weight:bold;" target="_blank" title="Click to know more about ISOTHERMAL">ISOTHERMAL</a>") = -nRT "ln" (V_(2))/(V_(<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>)/(40) xx 2 xx 300 xx "ln" (<a href="https://interviewquestions.tuteehub.com/tag/5-319454" style="font-weight:bold;" target="_blank" title="Click to know more about 5">5</a>)/(10) = <a href="https://interviewquestions.tuteehub.com/tag/100-263808" style="font-weight:bold;" target="_blank" title="Click to know more about 100">100</a>` cal</body></html> | |
| 51754. |
10 grams of steam has higher internal energy than 10 grams of ice. Comment. |
| Answer» <html><body><p></p>Solution :The kinetic energy of molecules in <a href="https://interviewquestions.tuteehub.com/tag/steam-1226863" style="font-weight:bold;" target="_blank" title="Click to know more about STEAM">STEAM</a> (<a href="https://interviewquestions.tuteehub.com/tag/gas-1003521" style="font-weight:bold;" target="_blank" title="Click to know more about GAS">GAS</a>) will be <a href="https://interviewquestions.tuteehub.com/tag/greater-476627" style="font-weight:bold;" target="_blank" title="Click to know more about GREATER">GREATER</a> than the kinetic energy of molecules in icc (<a href="https://interviewquestions.tuteehub.com/tag/solid-1216587" style="font-weight:bold;" target="_blank" title="Click to know more about SOLID">SOLID</a>). Hence internal energy of stcam is greater than that of ice.</body></html> | |
| 51755. |
10 grams of a sample of potassium chlorate gave on complete decomposition 2.24 litre of oxygen at N.T.P. What is the percentage purity of the sample ? |
| Answer» <html><body><p><br/></p>Solution :Chemical <a href="https://interviewquestions.tuteehub.com/tag/equation-974081" style="font-weight:bold;" target="_blank" title="Click to know more about EQUATION">EQUATION</a> for the reaction is : <br/> `underset(underset(=245g)(2xx(39+35.5+48)))(2KClO_(3)(s))overset("Heat")(<a href="https://interviewquestions.tuteehub.com/tag/rarr-1175461" style="font-weight:bold;" target="_blank" title="Click to know more about RARR">RARR</a>)2KCl(s)+underset(underset(=67.2L)(3xx22.4))(3O_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)(<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/> 67.2 L of `O_(2)` is evolved at N.T.P from `KClO_(3) 245 g` <br/> 2.24 L of `O_(2)` is evolved at N.T.P from `KClO_(3)=(245)/(67.5)xx2.24=8.167g` <br/> Percentage purity of sample `= ("Mass of pure "KClO_(3))/("Mass of the sample")xx100=(8.167)/(10.0)xx100=81.7%`.</body></html> | |
| 51756. |
10 grams of a mixture of CaCO_(3) and MgCO_(3) require equal weight of sulphuric acid for complete reaction. Calculate the percentage weight of MgCO(3) in the mixture. |
| 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> :0.11</body></html> | |
| 51757. |
10 grams of a hydrated sodium carbonate, Na_(2)CO_(3).xH_(2)O, on strong heating loses a weight of 6.29 grams. Report the value of x. |
| 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/hydrated-2721634" style="font-weight:bold;" target="_blank" title="Click to know more about HYDRATED">HYDRATED</a> sodium carbonate <a href="https://interviewquestions.tuteehub.com/tag/loses-1079374" style="font-weight:bold;" target="_blank" title="Click to know more about LOSES">LOSES</a> water of <a href="https://interviewquestions.tuteehub.com/tag/hydration-1033911" style="font-weight:bold;" target="_blank" title="Click to know more about HYDRATION">HYDRATION</a> on strong heating. <br/> `Na_(2)CO_(3).xH_(2)Orarr Na_(2)CO_(3)+xH_(2)O` <br/> 1 <a href="https://interviewquestions.tuteehub.com/tag/mole-1100299" style="font-weight:bold;" target="_blank" title="Click to know more about MOLE">MOLE</a> of `Na_(2)CO_(3)-=X` moles of `H_(2)O` <br/> 106 grams of `Na_(2)CO_(3)-=X xx18` grams of water <br/> 3.71 grams of `Na_(2)CO_(3)-=6.29` grams of water <br/> 3.71 grams of `Na_(2)CO_(3)` is related to 106 grams <br/> 6.29 grams of water is related to ? <br/> Weight of water in one formula unit `=(6.29)/(3.71)xx106`<br/> = 180 grams <br/> 18 grams of water = 1 mole <br/> 180 grams of water = ? <br/> Value of `X=(180)/(18)=10`</body></html> | |
| 51758. |
10 grams of 90% pure limestone is thermally decomposed. If the gas liberated is utilised for the conversion of aqueous sodium carbonate, how many moles of bicarbonate is formed? |
| 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> :0.18 <a href="https://interviewquestions.tuteehub.com/tag/mol-1100196" style="font-weight:bold;" target="_blank" title="Click to know more about MOL">MOL</a></body></html> | |
| 51759. |
1.0 gallon pure octane ( "density" 2.65 kg//"gallon") on combustion prodeces 11.53 kg CO,CO_(2) and H_(2)O.CO is formed partially due to combustion of octane which is responsible to decrease the efficiency of engine. If complete combustion of octane to CO_(2) and H_(2)O provide 100% efficiency to engine, calculate efficiency of engine in the above case. |
| 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> :`95.5%`;</body></html> | |
| 51760. |
1.0 g sample of the Rochelle salt [NaOOC-CHOH-CHOH-COOK] (NaKC_(4)H_(4)O_(6).4H_(2)O)(Mw=282), on ignition, is converted into NaKCO_(3)(Mw=122), which is titrated with 50 " mL of " 0.1 MH_(2)SO_(4). The excess of H_(2)SO_(4) requries 30 mL 0.2 M KOH. |
| Answer» <html><body><p></p>Solution :`NaKC_(4)H_(4)O_(6).4H_(2)Ooverset(Delta)toNaK.CO_(3)+3CO_(2)uarr+H_(2)Ouarr` <br/> Total m" Eq of "`H_(2)SO_(4)=50xx0.1xx2` (n factor) `=10.0` <br/> <a href="https://interviewquestions.tuteehub.com/tag/excess-978535" style="font-weight:bold;" target="_blank" title="Click to know more about EXCESS">EXCESS</a> m" Eq of "`H_(2)SO_(4)=m" Eq of "<a href="https://interviewquestions.tuteehub.com/tag/naoh-572531" style="font-weight:bold;" target="_blank" title="Click to know more about NAOH">NAOH</a> used` <br/> `=30xx0.2xx1` (n factor) <br/> `=6.0` <br/> m" Eq of "`H_(2)SO_(4)` used for `NaKCO_(3)` <br/> `=m" Eq of "H_(2)SO_(4)` (total)`-m" Eq of "H_(2)SO_(4)` used for NaOH <br/> `=10-6=4.0` <br/> `thereforem" Eq of "NaKCO_(3)=4.0` <br/> `implies` n-factor of `NaKCO_(3)=2`, during its neutralisation with <br/> `H_(2)SO_(4)(" Eq of "NaKCO_(3)=(<a href="https://interviewquestions.tuteehub.com/tag/122-271030" style="font-weight:bold;" target="_blank" title="Click to know more about 122">122</a>)/(2))` <br/> Weight of `NaKCO_(3)=4.0xx10^(-3)<a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a>(122)/(2)=0.244g` <br/> 1 " <a href="https://interviewquestions.tuteehub.com/tag/mol-1100196" style="font-weight:bold;" target="_blank" title="Click to know more about MOL">MOL</a> of "Rochelle salt`=1 " mol of "NaKCO_(3)` <br/> Therefore, weight of rochelle salt`=(282xx0.244)/(122)=0.364g` <br/> `%` purity of Rochelle salt`=(0.364xx100)/(1.0)=36.4%`</body></html> | |
| 51761. |
10 g sample of H_(2)O_(2) just decolorised 100 ml of 0.1 M KMnO_(4) in acidic medium % by mass of H_(2)O_(2) in the sample is |
| Answer» <html><body><p>`3.40`<br/>`8.5`<br/>`17.0`<br/>`1.70`</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/eq-446394" style="font-weight:bold;" target="_blank" title="Click to know more about EQ">EQ</a>. `H_(2)O_(2)` = Eq. `KMnO_(4)` <br/> `(w)/(34)xx2=0.1xx0.1xx5impliesw=0.85gm` <br/> implies % purity = `(0.85)/(<a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a>)xx100=8.5%`</body></html> | |
| 51762. |
10 g of sample of mixture of CaCl_(2) and NaCl are treated to precipitate all the calcium as CaCO_(3). This CaCO_(3) is heated to convert all the Ca to CaO and the final mass of CaCl_(2) in the original mixture is |
| Answer» <html><body><p>0.321<br/>0.162<br/>0.218<br/>0.12<br/></p>Solution :a) `CaCl_(2) + NaCl = 10g` <br/> Let, weight of `CaCl_(2)` = <a href="https://interviewquestions.tuteehub.com/tag/xg-747279" style="font-weight:bold;" target="_blank" title="Click to know more about XG">XG</a> <br/> `CaCl_(2) to CaCO_(3) to CaO` <br/> Moleof CaO = `(1.62)/(56) rArr x/111 = 1.62/56 x = 3.21g` <br/> `%` of `CaCl_(2)` = (3.21/10 x 100) = 32.1%`</body></html> | |
| 51763. |
10 g of sucrose are dissoled in 100 g of water. Find the mass percentage of sucrose in the solution. |
| Answer» <html><body><p></p>Solution :<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 sucrose <a href="https://interviewquestions.tuteehub.com/tag/dissolved-956358" style="font-weight:bold;" target="_blank" title="Click to know more about DISSOLVED">DISSOLVED</a> = 10 <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/> Mass 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> (solvent) = 100 g <br/> `therefore` Total mass of the solution `=100 + 10 = 110 g` <br/> <a href="https://interviewquestions.tuteehub.com/tag/hence-484344" style="font-weight:bold;" target="_blank" title="Click to know more about HENCE">HENCE</a>, mass percentage of sucrose <br/> `=("Mass of solute")/("Mass of solution")xx 100 = 10/110 xx 100 = 9.09`</body></html> | |
| 51764. |
1.0 g of Mg is burnt in a closed vessel which contains 0.5 g of O_2. (i) Which reactant is left in excess? (ii) Find the weight of excess reactant. (iii) How many millilitres of 0.5 N H_2 SO_4 will dissolve the residue in the vessel? |
| Answer» <html><body><p></p>Solution :`Mg+O_2 to MgO` <br/> Equivalent of `Mg = (1)/(12) = 0.0833""("eq. wt. of" Mg = ( "at. wt." )/( "<a href="https://interviewquestions.tuteehub.com/tag/valency-724638" style="font-weight:bold;" target="_blank" title="Click to know more about VALENCY">VALENCY</a>") = ( <a href="https://interviewquestions.tuteehub.com/tag/24-295400" style="font-weight:bold;" target="_blank" title="Click to know more about 24">24</a>)/(2) )` <br/> Equivalent of oxygen `(0.5)/( 8) = 0.0625 "(eq. wt. of O = 8)"` <br/> (i) Mg is in excess because its eq. is <a href="https://interviewquestions.tuteehub.com/tag/greater-476627" style="font-weight:bold;" target="_blank" title="Click to know more about GREATER">GREATER</a> than that of oxygen. <br/> (ii) Equivalent of Mg in excess `=0.0833 - 0.0625` <br/> `= 0.0208` <br/> `therefore` weight of Mg in excess `=` equivalent `<a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a>` equivalent weight <br/> `= 0.0208 xx 12 = 0.25 g` <br/> (iii) The residue contains MgO and Mg which has not taken part in the reaction. Suppose `upsilon` mL of `H_2 SO_4` is required to <a href="https://interviewquestions.tuteehub.com/tag/dissolve-432057" style="font-weight:bold;" target="_blank" title="Click to know more about DISSOLVE">DISSOLVE</a> the residue. <br/> `therefore` m.e. of `H_2 SO_4` = m.e. of Mg + m.e. of MgO (not reacted) <br/> `therefore 0.5 upsilon` = (eq. of Mg + eq. of MgO)<br/> ` = 1000` (eq. of Mg `+` eq. of O) <br/> `=1000 (0.0208 + 0.0625)` <br/> `= 83.3` <br/> `therefore upsilon= 166.6` mL.</body></html> | |
| 51765. |
1.0 g of metal carbonate neutralises 200 mL of 0.1 N HCl. The equivalent mass of the metal will be : |
| Answer» <html><body><p>50<br/>40<br/>20<br/>100</p>Solution :N//A</body></html> | |
| 51766. |
1.0 g of magnesium is burnt with 0.56 g of O_(2) in a closed vessel. Which reactant is left in excess and how much ? |
| Answer» <html><body><p>Mg , 0.44 g<br/>`O_(2)` , 0.28 g<br/>Mg , 0.16 g<br/>`O_(2)` , 0.16 g</p>Solution :`underset(2xx24g)(2Mg)+underset(<a href="https://interviewquestions.tuteehub.com/tag/32g-307750" style="font-weight:bold;" target="_blank" title="Click to know more about 32G">32G</a>)(O_(2))rarr2MgO` <br/> 32 g of `O_(2)` react with Mg = <a href="https://interviewquestions.tuteehub.com/tag/48-318016" style="font-weight:bold;" target="_blank" title="Click to know more about 48">48</a> g <br/> 0.56 g of `O_(2)` react with `Mg=((<a href="https://interviewquestions.tuteehub.com/tag/48g-1880255" style="font-weight:bold;" target="_blank" title="Click to know more about 48G">48G</a>))/((32g))xx(0.56g)` <br/> Mass of mg <a href="https://interviewquestions.tuteehub.com/tag/taken-659096" style="font-weight:bold;" target="_blank" title="Click to know more about TAKEN">TAKEN</a> `= 1.0 g` <br/> Mass of Mg <a href="https://interviewquestions.tuteehub.com/tag/left-1070879" style="font-weight:bold;" target="_blank" title="Click to know more about LEFT">LEFT</a> `= (1- 0.84) = 0.16 g`.</body></html> | |
| 51767. |
1.0 g of magnesium is burnt with 0.56 g O_(2) in a closed vessel. Which reaction is left in excess and how much ? ( At. wt. Mg = 24, O = 16) |
| Answer» <html><body><p>`<a href="https://interviewquestions.tuteehub.com/tag/mg-1095425" style="font-weight:bold;" target="_blank" title="Click to know more about MG">MG</a>, 0.16 g `<br/>`O_(2), 0.16 g`<br/>`Mg, 0.16 g`<br/>`O_(2), 0.28 g`</p>Solution :`M g + 1/2 O_(2) <a href="https://interviewquestions.tuteehub.com/tag/rarr-1175461" style="font-weight:bold;" target="_blank" title="Click to know more about RARR">RARR</a> MgO` <br/> `(1.0)/(<a href="https://interviewquestions.tuteehub.com/tag/24-295400" style="font-weight:bold;" target="_blank" title="Click to know more about 24">24</a>) (0.56)/(<a href="https://interviewquestions.tuteehub.com/tag/32-307188" style="font-weight:bold;" target="_blank" title="Click to know more about 32">32</a>)` <br/> Oxygen is <a href="https://interviewquestions.tuteehub.com/tag/limiting-2791961" style="font-weight:bold;" target="_blank" title="Click to know more about LIMITING">LIMITING</a> reagent so `(0.5)/(12) - x(0.07)/(4) - x/2` <br/> `(0.07)/(4) - x/2 = 0` <br/> `x=(0.07)/(2)` <br/> excess `Mg=(0.5)/(12) - (0.07)/(2)` mole <br/> `rArr 1-0.7xx12= 0.16g`</body></html> | |
| 51768. |
1.0 g of magnesium is burnt with 0.56 g O_2 in a closed vessel. Which reactant is left in excess and how much? (At.wt: Mg=24, O = 16) |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/mg-1095425" style="font-weight:bold;" target="_blank" title="Click to know more about MG">MG</a>, 0.16 <a href="https://interviewquestions.tuteehub.com/tag/g-1003017" style="font-weight:bold;" target="_blank" title="Click to know more about G">G</a><br/>`O_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>), 0.16 g`<br/>Mg, 0.44 g<br/>`O_(2), 0.28 g` </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> | |
| 51769. |
10 g of hydrogen and 64 g of oxygen were filled in a steel vessel and exploded. The amount of water produced in this reaction will be |
| Answer» <html><body><p>3 mole<br/><a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a> mole<br/>2 mole<br/>1 mole</p>Solution :10 g of `H_(2)+ 64g` of `O_(2) to` <a href="https://interviewquestions.tuteehub.com/tag/water-1449333" style="font-weight:bold;" target="_blank" title="Click to know more about WATER">WATER</a> <br/>`2H_(2) + O_(2) to 2H_(2)O` <br/>4 g of hydrogen react with 32 g of oxygen. So 10 g of hydrogen will react with <a href="https://interviewquestions.tuteehub.com/tag/80g-338891" style="font-weight:bold;" target="_blank" title="Click to know more about 80G">80G</a> of oxygen. But we have <a href="https://interviewquestions.tuteehub.com/tag/given-473447" style="font-weight:bold;" target="_blank" title="Click to know more about GIVEN">GIVEN</a> amount of oxygen only 64 g. It means, here oxygen is the limiting agent. Now all the oxygen react with 8 g of hydrogen and form 4 moles of water.</body></html> | |
| 51770. |
10 g of hydrogen and 64 g of oxygen were filled in a steel vessel and exploded. Amount of water produced in the reaction will be : |
| Answer» <html><body><p>3 mol<br/>4 mol<br/>1 mol<br/>2 mol</p>Solution :`underset("1 mol")(H_(2)(g))+underset("0.5 mol")(1//2O_(2)(g))rarrunderset("1 mol")(H_(2)O(l))` <br/> 10 gof `H_(2)=((<a href="https://interviewquestions.tuteehub.com/tag/10g-267074" style="font-weight:bold;" target="_blank" title="Click to know more about 10G">10G</a>))/(("2 g mol"^(-1)))=5` mol <br/> 64 g of `O_(2)=((64g))/(("<a href="https://interviewquestions.tuteehub.com/tag/32-307188" style="font-weight:bold;" target="_blank" title="Click to know more about 32">32</a> g mol"^(-1)))=2` mol <br/> In this reaction, `O_(2)` is <a href="https://interviewquestions.tuteehub.com/tag/limiting-2791961" style="font-weight:bold;" target="_blank" title="Click to know more about LIMITING">LIMITING</a> reactant 0.5 mol of <a href="https://interviewquestions.tuteehub.com/tag/oxygen-1144542" style="font-weight:bold;" target="_blank" title="Click to know more about OXYGEN">OXYGEN</a> from water = 1 mol <br/> 2 mol of oxygen from water <br/> `= (("1 mol"))/(("0.5 mol"))xx("2 mol")=4` mol.</body></html> | |
| 51771. |
10 g of hydrofluoric acid gas occupies 5.6 lit of volume at STP . If the empirical formula of the gas is HF , then its molecular formula in the gaseous state will be |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/hf-479831" style="font-weight:bold;" target="_blank" title="Click to know more about HF">HF</a><br/>`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>)F_(2)`<br/>`H_(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)F_(3)`<br/>`H_(4)F_(4)`</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> | |
| 51772. |
1 g of Mg is burnt in a closed vessel containing 0.5 g of O_2. Which reactant is limiting reagent and how much of the excess reactant will be left? |
| Answer» <html><body><p></p>Solution :Carbon burns in oxygen according to the following <a href="https://interviewquestions.tuteehub.com/tag/equation-974081" style="font-weight:bold;" target="_blank" title="Click to know more about EQUATION">EQUATION</a>. <br/> `underset(12.01 <a href="https://interviewquestions.tuteehub.com/tag/g-1003017" style="font-weight:bold;" target="_blank" title="Click to know more about G">G</a>)C + underset(32.0 g)(O_(2)) to underset(12.01 + 32.0 = 44.01 g)(CO_(2))` <br/> `therefore` 12.01 g of carbon require for combustion, = 32.0 g <br/> `therefore 1.0` g of carbon will require `=(32.0)/(12.01) xx 1.0 = 2.7 g` <br/> Hence, the oxygen present in the closed vessel is much <a href="https://interviewquestions.tuteehub.com/tag/less-1071906" style="font-weight:bold;" target="_blank" title="Click to know more about LESS">LESS</a> than the required amount. The amount of `O_(2)`will thus be completely consumed and carbon will be left in excess. Thus, `O_2` is the <a href="https://interviewquestions.tuteehub.com/tag/limiting-2791961" style="font-weight:bold;" target="_blank" title="Click to know more about LIMITING">LIMITING</a> reagent in this case. <br/> `therefore 32.0` g of oxygen burn carbon = 12.01 g <br/> 1.5 g of oxygen will burn carbon `=12.01/32.0 xx 1.5 = 0.56 g` <br/> Hence, carbon left unreacted = `1.0 - 0.56 = 0.44` g Now, `therefore32.0` g of oxygen combine with carbon to <a href="https://interviewquestions.tuteehub.com/tag/give-468520" style="font-weight:bold;" target="_blank" title="Click to know more about GIVE">GIVE</a> `CO_2 = 44.01` g <br/> `therefore` 1.5 g of oxygen will give `CO_(2) = (44.01)/(32.0) xx 1.5 = 2.1 g` <br/> Hence, carbon is left in excess in the given reaction. The amount of excess reactant is 0.44 g and the mass of the product `CO_2` formed in the reaction is 2.1 g</body></html> | |
| 51773. |
10 g of argonis compressd isothermally and reversibly at a temperature of27^(@)C from10L to 5L . Calculate q,w,DeltaE and DeltaHfor this process. R =2.0 cal K^(-1)mol^(-1) , log= 2 = 0.30. Atomic wt. of Ar=40 |
| Answer» <html><body><p></p>Solution :`q= 2.303 nRT log. (V_(2))/(V_(1)) = 2.303 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>)/( 40) xx 300 xx log. (5)/(10)= -103.635 cal`<br/> For <a href="https://interviewquestions.tuteehub.com/tag/isothermal-520014" style="font-weight:bold;" target="_blank" title="Click to know more about ISOTHERMAL">ISOTHERMAL</a> expansion,`DeltaE= <a href="https://interviewquestions.tuteehub.com/tag/0-251616" style="font-weight:bold;" target="_blank" title="Click to know more about 0">0</a>``w= DeltaE-q= 0-( - 103.635)= +103.635 cal` <br/>Also, when temperature is constant, `P_(1)V_(1) = P_(2)V_(2)` or `PV =` constant `DeltaH = DeltaE+Delta(PV) = 0+0=0`</body></html> | |
| 51774. |
10 grams of CaCO_(3) is completely decomposed to x and CaO. 'x' is passed into an aqueous solution containing 0.1mole of sodium carbonate. What is the number of moles of sodium bicarbonate formed? (mol. wts: CaCO_(3) = 100, NaCO_(3) = 106, NaHCO_(3) = 84) |
| 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/>`0.1`<br/>`0.01`<br/>10</p>Solution :`underset(0.1)(CaCO_(3))rarrCaO+underset(0.1)underset("X")(CO_(2))` <br/> `underset(0.1)(CO_(2))+Na_(2)CO_(3)+H_(2)Orarrunderset(0.2)(2NaHCO_(3))`</body></html> | |
| 51775. |
100 g of a pure sample of CaCO_3 is treated a with 500 cm^(3) of M/2HCI solution. Calculate the of CO_2 that will be evolved at S.T.P. |
| Answer» <html><body><p></p>Solution :The corresponding chemical <a href="https://interviewquestions.tuteehub.com/tag/equation-974081" style="font-weight:bold;" target="_blank" title="Click to know more about EQUATION">EQUATION</a> is: <br/> `underset(40.08 +12.01 + 48.0=100.1)(CaCO_(3)) + underset(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a> xx 36.46= 72.92)(2HCl) to CaCl_(2) + H_(2)O + underset("22.4 <a href="https://interviewquestions.tuteehub.com/tag/l-535906" style="font-weight:bold;" target="_blank" title="Click to know more about L">L</a> at S.T.P")(CO_(2))` <br/> The mass of HCl present in `50.0 cm^(3)` of `M/2` of HCl <br/> `=(1/2 xx 36.46 xx 50.0)/1000 = 0.911 g` <br/> From the equation it is clear that 1 g of `CaCO_3` will require 0.73 g of <a href="https://interviewquestions.tuteehub.com/tag/hci-479434" style="font-weight:bold;" target="_blank" title="Click to know more about HCI">HCI</a>. Therefore, the mass of HCI used is more than the required mass. This implies that `CaCO_3` is the limiting reagent. <br/> `therefore 100.1 g` of `CaCO_(3)` give a volume of `CO_(2) = 22.4 L` <br/> `therefore 1.0 g` of `CaCO_(3)` give a volume of `CO_(2) = 22.4 L` <br/> `therefore 1.0 g` of `CaCO_(3)` will give a volume of `CO_(2)` <br/> `=22.4/100.1 xx 1.0 = 0.22 L` <br/> Hence,the volumeof carbondioxide evolved =0.22 L.</body></html> | |
| 51776. |
1.0 g of a mixture of carbonates of calcium and magnesium gave 240 cm^(3) of CO_(2) at N.T.P Calculate the percentage composition of the mixture. |
| Answer» <html><body><p><br/></p>Solution :Let the mass of `CaCO_(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)` in the <a href="https://interviewquestions.tuteehub.com/tag/mixture-1098735" style="font-weight:bold;" target="_blank" title="Click to know more about MIXTURE">MIXTURE</a> =x g <br/> `:.` The mass of `MgCO_(3)` in the mixture `= (1-x)g` <br/> Step I. Volume of `CO_(2)` evolved from `CaCO_(3)` <br/> `underset(40+12+48+100g)(CaCO_(3)(s))overset("Heat")(rarr)<a href="https://interviewquestions.tuteehub.com/tag/cao-411869" style="font-weight:bold;" target="_blank" title="Click to know more about CAO">CAO</a>(s)+underset(22400"cc")(CO_(2)(g))` <br/> 100 g of `CaCO_(3)` evolve `CO_(2)` upon heating = 22400 cc <br/> xg of `CaCO_(3)` evolve `CO_(2)` upon upon heating `= (22400)/(100) xx x c c = (224x)c c`. <br/> Step II. Volume of `CO_(2)` evolved from `MgCO_(3)` <br/> `underset(24+12=48=84g)(MgCO_(3)(s))overset("Heat")(rarr)MgO(s)+underset("22400cc")(CO_(2)(g))` <br/> 84 g of `MgCO_(3)` evolve `CO_(2)` upon heating = 22400 cc <br/> `(1-x)`g of `MgCO_(3)` evolve `CO_(2)` upon heating `= 22400 xx ((1-x))/(84) c c. = 266.6 xx (1-x)c c` <br/> Step III. Percentage composition of the mixture <br/> According to available data : <br/> `224 x + 266.6 (1-x)=240` <br/> `224 x + 266.6 - 266.6 x = 240` <br/> `- 42.6 x= -26.6 or x = (26.6)/(42.6)=0.625` <br/> `:.` Percentage of `CaCO_(3)` in the mixture `= 0.625 xx 100 = 62.5 %` <br/> Percentage of `MgCO_(3)` in the mixture `= 100 - 62.5 = 37.5 %`.</body></html> | |
| 51777. |
1.0 g NaHSO_3 and Na_2SO_3 was dissolved in water to prepare a 200 mL solution. Two separate experiments were carried out. (a). 25 " mL of " sample was mixed with 25 " mL of " I_2 solution and excess of I_2 left after the reaction with NaHSO_3 and Na_2SO_3 was back titrated with 0.1002 N Na_2S_2O_3, 1.34 " mL of " which was required (25 " mL of " I_2 solution is equivalent to 24.20" mL of " Na_2S_2O_3 solution). (b). 50 " mL of " sample was oxidised to Na_2SO_4 by the action of H_2O_2,H_2SO_4 sormed (from NaHSO_3) was titrated with 22.3 " mL of " 0.1 N NaOH. Find percentage of NaHSO_3 and Na_2SO_3 in the original sample. |
| Answer» <html><body><p></p>Solution :1 <a href="https://interviewquestions.tuteehub.com/tag/g-1003017" style="font-weight:bold;" target="_blank" title="Click to know more about G">G</a> mixture is present in 200 mL <br/> (a). <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 of " mixture] <br/> `impliesmEq` of `NaHSO_3+mEq` of `Na_2SO_3=mEq` of ` I_2` <br/> Used <br/> `mEq` of ` I_2` used`=24.20-1.34xx(0.1002)/(25)mL` <br/> `=22.905` <br/> m" Eq of "`I_2` in 200 mL`=(200)/(25)xx22.905=18.32` <br/> (b). 50 mL is oxidised to `Na_2SO_4` <br/> `H_2SO_4` formed by `NaHSO_3-=22.3mL` of `0.1` N `NaOH` <br/> `mEq` of `NaHSO_3=22.3xx0.1=(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>.23)/(25)mL` <br/> `mEq` of ` NaHSO_2 i n 200mL=2.23xx(200)/(50)=8.92` <br/> `[{:("Ew of "NaHSO_(3)" in experiment"," "),((i)=(Mw)/(2)," "),(underset(x=4)underset(1+x-6=-1)(HSO_(3)^(ɵ)),tounderset(x=6)underset(x-8=-2)(SO_(4)^(2-)(x=2))):}]` <br/> `[{:("Ew of "NaHSO_(3) " in experiment"," "),((ii)=(Mw)/(1)," "),(HSO_(3)^(ɵ),toH^(o+)+SO_(3)^(2-)):}]` <br/> M" Eq of "`Na_(2)SO_(3)` in the original solution <br/> `=18.32-2xx8.92=0.48` <br/> Weight of `Na_2SO_3=8.92mEqxx(1)/(1000)xxEw of "NaHSO_(3)` <br/> `["Ew of "NaHSO_(3)=(Mw)/(1)=<a href="https://interviewquestions.tuteehub.com/tag/104-266430" style="font-weight:bold;" target="_blank" title="Click to know more about 104">104</a>]` <br/> `=(8.92)/(1000)xx104=0.92` <br/> `therefore%` of `NaHSO_(3)=(0.927)/(1)xx100=92.7%` <br/> [<a href="https://interviewquestions.tuteehub.com/tag/equivalent-446407" style="font-weight:bold;" target="_blank" title="Click to know more about EQUIVALENT">EQUIVALENT</a> weight of `Na_(2)SO_(3)=(Mw)/(2)=(126)/(2)=63]` <br/> Weight of `Na_(2)SO_(3)=0.48mEqxx(1)/(1000)xx63=0.3g` <br/> `%` of `Na_(2)SO_(3)=(0.3)/(1)xx100=3%` <br/> `NaHSO_(3)=92.7%""Na_(2)SO_(3)=3%`</body></html> | |
| 51778. |
1.0 g carbonate of a metal was dissolved in 50 mL N//20 HCl solution. The resulting liquid required 25 mL of N//5 NaOH solution to neutralise it completely. Calculate the equivalent mass of the metal carbonate. |
| 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> | |
| 51779. |
10 cm^(3) of 0.1 N monobasic acid requires 15 cm^(3) of sodium hydroxide solution whose normality is : |
| Answer» <html><body><p>0.066 <a href="https://interviewquestions.tuteehub.com/tag/n-568463" style="font-weight:bold;" target="_blank" title="Click to know more about N">N</a><br/>0.66 N<br/><a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>.5 N<br/>0.15 N</p>Solution :`N_(1)V_(1)=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>)V_(2)` <br/> `0.1xx10=N_(2)xx15` <br/> `N_(2)=0.066`</body></html> | |
| 51780. |
10^(-6) M NaOH solution is diluted 100 times. Calculate the pH of the diluted base. |
| Answer» <html><body><p><br/></p>Solution :After dilution, `[<a href="https://interviewquestions.tuteehub.com/tag/oh-585115" style="font-weight:bold;" target="_blank" title="Click to know more about OH">OH</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>^(-<a href="https://interviewquestions.tuteehub.com/tag/8-336412" style="font-weight:bold;" target="_blank" title="Click to know more about 8">8</a>) M`. <a href="https://interviewquestions.tuteehub.com/tag/hence-484344" style="font-weight:bold;" target="_blank" title="Click to know more about HENCE">HENCE</a>, `[OH^(-)]` from `H_(2)O` cannot be neglected. <a href="https://interviewquestions.tuteehub.com/tag/proceed-592932" style="font-weight:bold;" target="_blank" title="Click to know more about PROCEED">PROCEED</a> as in Solved Problem 2 of Type VIII above.</body></html> | |
| 51781. |
10^(-6)MNaOH is diluted 100 times. The pH of the diluted base is |
| Answer» <html><body><p>between 5 and 6<br/> between 6 and 7<br/> between 10and <a href="https://interviewquestions.tuteehub.com/tag/11-267621" style="font-weight:bold;" target="_blank" title="Click to know more about 11">11</a><br/> between 7 and 8</p>Solution :`[OH^(-)]` after dilution `= (<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))/(<a href="https://interviewquestions.tuteehub.com/tag/100-263808" style="font-weight:bold;" target="_blank" title="Click to know more about 100">100</a>)=10^(-8)M` <br/> `:. OH^(-)` <a href="https://interviewquestions.tuteehub.com/tag/ions-1051295" style="font-weight:bold;" target="_blank" title="Click to know more about IONS">IONS</a> from `H_(2)O` cannot be neglected. <br/> Total `[OH^(-)]=10^(-8)+10^(-7)=10^(-8)xx11` <br/> `:. pOH=-log(11xx10^(-8))=8-1.04=6.96` <br/> `:. pH = 14 - 6.96 = 7.04`</body></html> | |
| 51782. |
10^(-6) is the prefix for ......... . |
| 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> :<a href="https://interviewquestions.tuteehub.com/tag/micro-1095491" style="font-weight:bold;" target="_blank" title="Click to know more about MICRO">MICRO</a></body></html> | |
| 51783. |
10^(-6) can be written as PREFIX : |
| Answer» <html><body><p>micro<br/>mili<br/>mega<br/>deca</p>Answer :A</body></html> | |
| 51784. |
10^(-4) g of gelatin is required to be added to 100 cm^(3) of a standard gold sol to just prevent its coagulation by the addition of 1 cm^(3) of 10% NaCl solution to it. Hence, the gold number of gelatin is |
| Answer» <html><body><p>10<br/>1<br/>0.1<br/>0.01</p>Solution :By definition, gold number is the number of milligrams of dry <a href="https://interviewquestions.tuteehub.com/tag/protective-1170744" style="font-weight:bold;" target="_blank" title="Click to know more about PROTECTIVE">PROTECTIVE</a> colloids required to just prevent the coagulation of 10mL of red gold sol when 1 <a href="https://interviewquestions.tuteehub.com/tag/ml-548251" style="font-weight:bold;" target="_blank" title="Click to know more about ML">ML</a> (or 1 `cm^(3)` ) of 10% NaCl solution is added to it.<br/> Here the amount of <a href="https://interviewquestions.tuteehub.com/tag/gelatin-471595" style="font-weight:bold;" target="_blank" title="Click to know more about GELATIN">GELATIN</a> required to protect 10 mL (or 10 `cm^(3)`) of gold sol <br/> `=(10^(-4)xx10)/(100)=10^(-<a href="https://interviewquestions.tuteehub.com/tag/5-319454" style="font-weight:bold;" target="_blank" title="Click to know more about 5">5</a>) g` <br/> `=10^(-5)xx10^(-3) mg=10^(-2)` mg <br/> =0.01 mg <br/> `therefore` Gold number =0.01</body></html> | |
| 51785. |
10^(-3) gm equivalents of K_(2) Cr_(2) O_(7)in 50% H_(2) SO_(4) is needed to oxidise all the organic matter present in 1 lit of water. Then COD of water 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> ppm<br/>10 ppm<br/>12 ppm<br/>8 ppm</p>Answer :D</body></html> | |
| 51786. |
1. When 22.4 litres of H_2(g) is mixed with 11.2 litres of Cl_2 (g) each at 273 K at 1 atm the moles of HCl (g), formed is equal to....... |
| Answer» <html><body><p> 2 <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 <a href="https://interviewquestions.tuteehub.com/tag/hcl-479502" style="font-weight:bold;" target="_blank" title="Click to know more about HCL">HCL</a> (<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/>0.5 moles of HCI (g) <br/><a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>.5 moles of HCl (g)<br/>1 mole of HCl (g)<br/></p>Solution :1 moleof HCI(g)<br/> Solutionn:` H_2 (g)+ Cl_2(g)to 2HCI (g)` <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/FM_CHE_XI_SP_20_E01_001_S01.png" width="80%"/></body></html> | |
| 51787. |
(1) What are auto redox reactions? Give example. (ii) Define orbital. What are the n and I values of 3p_x and 4d_(x^2-y^2) electron? |
| Answer» <html><body><p></p>Solution :Auto redox <a href="https://interviewquestions.tuteehub.com/tag/reactions-20919" style="font-weight:bold;" target="_blank" title="Click to know more about REACTIONS">REACTIONS</a> :Auto redor reactions. It is a rede reaction in which substance act as both the <a href="https://interviewquestions.tuteehub.com/tag/oxidizing-2906555" style="font-weight:bold;" target="_blank" title="Click to know more about OXIDIZING">OXIDIZING</a> agent and the <a href="https://interviewquestions.tuteehub.com/tag/reducing-2981618" style="font-weight:bold;" target="_blank" title="Click to know more about REDUCING">REDUCING</a> agent. (Reaction occur in the same substance) <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/FM_CHE_XI_SP_JUN_19_E01_034_S01.png" width="80%"/> <br/> (ii)1.Orbital is a three dimensional space which the probability of finding the electron is <a href="https://interviewquestions.tuteehub.com/tag/maximum-556915" style="font-weight:bold;" target="_blank" title="Click to know more about MAXIMUM">MAXIMUM</a><br/> 2. For `3p_x` eletron `""`n value =3 <br/> `"" "<a href="https://interviewquestions.tuteehub.com/tag/l-535906" style="font-weight:bold;" target="_blank" title="Click to know more about L">L</a> value" =1 `<br/> 3. For `4d_(x^2-y^2)` electron`""`n vlaue =4 <br/> `""` l value =2</body></html> | |
| 51788. |
1) Total number of electron paire= (1)/(2) (number of valence electrons pm electron (for ionic charge) 2) Number of bond electron pairs= number of atoms - 3) Number of electron pairs around central atom= total number of electron pairs - 3 [number atoms (except H)] 4) Number lone pair = (number of central electron pairs - number bond pairs) Read the above method and answer the following questions : Based on the given structure of the some of the molecules have been matched. Which is the incorrect matching ? |
| Answer» <html><body><p>`PCI_(5)^(-)` trigonal bipyramidal <br/>`CIO_(<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/square-650836" style="font-weight:bold;" target="_blank" title="Click to know more about SQUARE">SQUARE</a> <a href="https://interviewquestions.tuteehub.com/tag/planar-1155488" style="font-weight:bold;" target="_blank" title="Click to know more about PLANAR">PLANAR</a> <br/>`ICI_(4)^(-)` square planar <br/>`PCI_(4)^(+)` <a href="https://interviewquestions.tuteehub.com/tag/tetrahedral-2294863" style="font-weight:bold;" target="_blank" title="Click to know more about TETRAHEDRAL">TETRAHEDRAL</a> </p>Answer :B</body></html> | |
| 51789. |
1) Total number of electron paire= (1)/(2) (number of valence electrons pm electron (for ionic charge) 2) Number of bond electron pairs= number of atoms - 3) Number of electron pairs around central atom= total number of electron pairs - 3 [number atoms (except H)] 4) Number lone pair = (number of central electron pairs - number bond pairs) Read the above method and answer the following questions : Square planar shape is predicted for: |
| Answer» <html><body><p>`ICI_(<a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>)^(-), CIO_(3)^(-)` <br/>`PCI_(4)^(-),PCI_(6)^(-)` <br/>`ICI_(4)^(-),PCI_(4)^(+)` <br/>`ICI_(4)^(-),XeF_(4)`</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> | |
| 51790. |
1) Total number of electron paire= (1)/(2) (number of valence electrons pm electron (for ionic charge) 2) Number of bond electron pairs= number of atoms - 3) Number of electron pairs around central atom= total number of electron pairs - 3 [number atoms (except H)] 4) Number lone pair = (number of central electron pairs - number bond pairs) Read the above method and answer the following questions : Pair of species with same shape and same state of hybridisation of the central atom is : |
| Answer» <html><body><p>`PCI_(<a href="https://interviewquestions.tuteehub.com/tag/5-319454" style="font-weight:bold;" target="_blank" title="Click to know more about 5">5</a>) , ICI_(<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/>`NH_(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>), H_(2)O`<br/>`NH_(3),CIO_(3)^(-)`<br/>`ICI_(4)^(-),CIO_(3)^(-)`</p>Answer :C</body></html> | |
| 51791. |
1) The -=C-H unit of an alkyne is more acidic than a C – H unit of an alkene or alkane, allowingacetylene and terminal alkynes to be converted to their conjugate bases -= C. 2) Unlike alkenes, alkynes are reduced by metals, especially Li, Na and K. 3) Unlike alkenes, alkynes can undergo nucleophilic as well as electrophilic addition. Which of the following best describes what happens in the first step in the mechanism of the reactionshown? CH_3 CH_2CH_2CHBr_2 + 3NaNH_2 overset(NH_3)to CH_3CH_2C -=CNa +2NaBr +3NH_3 |
| Answer» <html><body><p><img src="https://doubtnut-static.s.llnwi.net/static/physics_images/AKS_ELT_AO_CHE_XI_V01_D_C04_E01_107_O01.png" width="30%"/> <br/><img src="https://doubtnut-static.s.llnwi.net/static/physics_images/AKS_ELT_AO_CHE_XI_V01_D_C04_E01_107_O02.png" width="30%"/> <br/><img src="https://doubtnut-static.s.llnwi.net/static/physics_images/AKS_ELT_AO_CHE_XI_V01_D_C04_E01_107_O03.png" width="30%"/> <br/><img src="https://doubtnut-static.s.llnwi.net/static/physics_images/AKS_ELT_AO_CHE_XI_V01_D_C04_E01_107_O04.png" width="30%"/> </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/conceptual-928163" style="font-weight:bold;" target="_blank" title="Click to know more about CONCEPTUAL">CONCEPTUAL</a></body></html> | |
| 51792. |
1) The -=C-H unit of an alkyne is more acidic than a C – H unit of an alkene or alkane, allowingacetylene and terminal alkynes to be converted to their conjugate bases -= C. 2) Unlike alkenes, alkynes are reduced by metals, especially Li, Na and K. 3) Unlike alkenes, alkynes can undergo nucleophilic as well as electrophilic addition. Which of the following best describes what happens in the first step in the mechanism of the hydrogen- deuterium exchange reaction shown? CH_3 -= CD underset(NH_3)overset(NaNH_2)to CH_3C -= CH |
| Answer» <html><body><p><img src="https://doubtnut-static.s.llnwi.net/static/physics_images/AKS_ELT_AO_CHE_XI_V01_D_C04_E01_106_O01.png" width="30%"/><br/><img src="https://doubtnut-static.s.llnwi.net/static/physics_images/AKS_ELT_AO_CHE_XI_V01_D_C04_E01_106_O02.png" width="30%"/><br/><img src="https://doubtnut-static.s.llnwi.net/static/physics_images/AKS_ELT_AO_CHE_XI_V01_D_C04_E01_106_O03.png" width="30%"/><br/><img src="https://doubtnut-static.s.llnwi.net/static/physics_images/AKS_ELT_AO_CHE_XI_V01_D_C04_E01_106_O04.png" width="30%"/></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/conceptual-928163" style="font-weight:bold;" target="_blank" title="Click to know more about CONCEPTUAL">CONCEPTUAL</a></body></html> | |
| 51793. |
1-Propanol and 2-propanol can be best distinguished by |
| Answer» <html><body><p>Oxidation with `KMnO_(<a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>)` follwed by reaction with Fehling solution<br/>Oxidation with acidic dichromate <a href="https://interviewquestions.tuteehub.com/tag/followed-2079285" style="font-weight:bold;" target="_blank" title="Click to know more about FOLLOWED">FOLLOWED</a> by reaction with Fehling solution<br/>Oxidation by heating with copper followed by reaction with Fehling solution<br/>Oxidation with concentrated `H_(2)SO_(4)` followed by reaction with Fehling solution.</p>Solution :1-Propanol `underset(-H_(2))overset(Cu//573K)rarr` Propanol <br/> (Give positive <a href="https://interviewquestions.tuteehub.com/tag/test-13460" style="font-weight:bold;" target="_blank" title="Click to know more about TEST">TEST</a> with Fehling solution) <br/> `2-"Propanol" underset(-H_(2))overset(Cu//573K)rarr "Propanone"` <br/> Oxidation with `KMnO_(4)` or `K_(2)Cr_(2)O_(7)/H^(+)` will give propanoic acid in case of 1-propanol which can not be tested with Fehlingsolution. Reaction with conc. `H_(2)SO_(4)` will <a href="https://interviewquestions.tuteehub.com/tag/lead-540361" style="font-weight:bold;" target="_blank" title="Click to know more about LEAD">LEAD</a> to dehydration and as such no oxidation is possible</body></html> | |
| 51794. |
1-Propanol and 2-Propanol can be best distinguished by : |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/oxidation-588780" style="font-weight:bold;" target="_blank" title="Click to know more about OXIDATION">OXIDATION</a> with alkaline `KMnO_(<a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>)` followed by reaction with Fehling solution<br/>oxidation with acidic dichromate followed by reaction with Fehling solution<br/>oxidation by heating with copper followed by reaction with Fehling solution<br/>oxidation with <a href="https://interviewquestions.tuteehub.com/tag/concentrated-409727" style="font-weight:bold;" target="_blank" title="Click to know more about CONCENTRATED">CONCENTRATED</a> `H_(2)SO_(4)` followed by reaction with Fehling solution</p>Answer :C</body></html> | |
| 51795. |
1-Phenylethanol can be prepared by reaction of benzaldehyde with |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/methyl-1095247" style="font-weight:bold;" target="_blank" title="Click to know more about METHYL">METHYL</a> iodide and Magnesium<br/>Methyl bromide<br/>Methyl bromide and `AlBr_(3)`<br/>`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/5-319454" style="font-weight:bold;" target="_blank" title="Click to know more about 5">5</a>)I` and `<a href="https://interviewquestions.tuteehub.com/tag/mg-1095425" style="font-weight:bold;" target="_blank" title="Click to know more about MG">MG</a>`</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> :`C_(6)H_(5)-overset(H)overset(|)(C)=O+CH_(3)MgI rarr C_(6)H_(5)-underset(CH_(3))underset(|)overset(H)overset(|)(C)-OMgI overset(H_(2)O //H^(+))rarr underset("1-Phenylethanol")(C_(6)H_(5)-underset(CH_(3))underset(|)overset(H)overset(|)(C)-OH)`</body></html> | |
| 51796. |
1) N_2+2O_2 |
| Answer» <html><body><p>`NO_2`<br/> `NO`<br/>`N_2O_5`<br/>`N_2O`</p>Solution :For the <a href="https://interviewquestions.tuteehub.com/tag/decomposition-946042" style="font-weight:bold;" target="_blank" title="Click to know more about DECOMPOSITION">DECOMPOSITION</a> of <a href="https://interviewquestions.tuteehub.com/tag/oxide-1144484" style="font-weight:bold;" target="_blank" title="Click to know more about OXIDE">OXIDE</a> of nitrogen, lower the <a href="https://interviewquestions.tuteehub.com/tag/equilibrium-974342" style="font-weight:bold;" target="_blank" title="Click to know more about EQUILIBRIUM">EQUILIBRIUM</a> constant <a href="https://interviewquestions.tuteehub.com/tag/greater-476627" style="font-weight:bold;" target="_blank" title="Click to know more about GREATER">GREATER</a> the stability of oxide.</body></html> | |
| 51797. |
1 mole SO_(2)=.......... |
| Answer» <html><body><p>`6.4g SO_(2)`<br/>`6.02xx10^(23)SO_(2)` molecule<br/>`22.4L SO_(2)` (<a href="https://interviewquestions.tuteehub.com/tag/ntp-572134" style="font-weight:bold;" target="_blank" title="Click to know more about NTP">NTP</a>)<br/>All of the above</p>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>::<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> | |
| 51798. |
1 molephotonhavingfrequency4.0 xx 10^(14) Hzfindenergy |
| 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> :159.0 <a href="https://interviewquestions.tuteehub.com/tag/kj-1063034" style="font-weight:bold;" target="_blank" title="Click to know more about KJ">KJ</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/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>)`</body></html> | |
| 51799. |
1 mole of water vapour is condensed to liquid at 25^(@)C. Now this water contains i) 3 moles of atoms ii) 1 mole of hydrogen molecules iii) 10 moles of electrons iv) 16 g of oxygen The correct combination is |
| Answer» <html><body><p>(i) & (ii) are <a href="https://interviewquestions.tuteehub.com/tag/correct-409949" style="font-weight:bold;" target="_blank" title="Click to know more about CORRECT">CORRECT</a> <br/>(i) & (<a href="https://interviewquestions.tuteehub.com/tag/iii-497983" style="font-weight:bold;" target="_blank" title="Click to know more about III">III</a>) are correct <br/>(i) & (<a href="https://interviewquestions.tuteehub.com/tag/iv-501699" style="font-weight:bold;" target="_blank" title="Click to know more about IV">IV</a>) are correct <br/>All are correct </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> | |
| 51800. |
1 mole of photons, each of frequency 250 s^(-1) would have approximately a total energy of _____ erg. |
| Answer» <html><body><p><br/></p>Solution :`E = <a href="https://interviewquestions.tuteehub.com/tag/nh-570695" style="font-weight:bold;" target="_blank" title="Click to know more about NH">NH</a> vartheta = 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> <a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a>^(23) xx 6.6 xx 10^(-27) xx 250 = <a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a> ` erg</body></html> | |