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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.
| 51951. |
0.5 mole each of two ideal gases A(C_(v,m)=(5)/(2)R) and B(C_(v,m)=3R) are taken in a container and expanded reversibely and adibatically , during this process, temperature of gaseous mixture decreased from 350 K and 250 K. Find DeltaH (in "cal"//"mol") for the process : |
| Answer» <html><body><p>`-100R`<br/>`-137.5R`<br/>`-<a href="https://interviewquestions.tuteehub.com/tag/375-309985" style="font-weight:bold;" target="_blank" title="Click to know more about 375">375</a> <a href="https://interviewquestions.tuteehub.com/tag/r-611811" style="font-weight:bold;" target="_blank" title="Click to know more about R">R</a>`<br/><a href="https://interviewquestions.tuteehub.com/tag/none-580659" style="font-weight:bold;" target="_blank" title="Click to know more about NONE">NONE</a> of these</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> | |
| 51952. |
0.5 mole each of H_2SO_4and CH_4 are kept in a container. A hole was made in the container. After 3 hours the order of partial pressures in the container will be: |
| Answer» <html><body><p><<a href="https://interviewquestions.tuteehub.com/tag/p-588962" style="font-weight:bold;" target="_blank" title="Click to know more about P">P</a>>`P_(SO_2) gt P_(CH_4) gt P_(H_2)`<br/>`P_(CH_4) gt P_(SO_2) gt P_(H_2)`<br/>`P_(H_2) gt P_(SO_2) gt P_(CH_4)`<br/>`P_(H_2) gt P_(CH_4) gt P_(SO_2)`<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> :`H_2 ` is <a href="https://interviewquestions.tuteehub.com/tag/lost-537630" style="font-weight:bold;" target="_blank" title="Click to know more about LOST">LOST</a> <a href="https://interviewquestions.tuteehub.com/tag/faster-985028" style="font-weight:bold;" target="_blank" title="Click to know more about FASTER">FASTER</a> `:. P_(H_2)` is <a href="https://interviewquestions.tuteehub.com/tag/least-7256596" style="font-weight:bold;" target="_blank" title="Click to know more about LEAST">LEAST</a> .</body></html> | |
| 51953. |
0.5 mol CaCO_3 solid decompose in 500 mL heated in closed vessel at 400 K reaction CaCO_(3(s)) hArr CaO_((s)) + CO_(2(g)) equilibrium constant of K_c = 0.9 "mol L"^(-1). Calculate mol of CO_2 at equilibrium how much percentage of reaction completed ? |
| 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> :`[CO_2]= 0.45 "<a href="https://interviewquestions.tuteehub.com/tag/mol-1100196" style="font-weight:bold;" target="_blank" title="Click to know more about MOL">MOL</a> L"^(-<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>)`, 90% decomposition <br/>`[CO_2]`= 0.3 mol and <a href="https://interviewquestions.tuteehub.com/tag/60-328817" style="font-weight:bold;" target="_blank" title="Click to know more about 60">60</a>% <a href="https://interviewquestions.tuteehub.com/tag/reaction-22747" style="font-weight:bold;" target="_blank" title="Click to know more about REACTION">REACTION</a> if `CaCO_3` and CaO used in calculation</body></html> | |
| 51954. |
0.5 litres of an ideal gas is present at a pressure of 700 mm of Hg and 30^@C. Find its volume at standard temperature and pressure. |
| Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :0.415 It</body></html> | |
| 51955. |
0.5 litres of an ideal gas is present at a pressure. |
| 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> :`0.415 I t`</body></html> | |
| 51956. |
0.5 g of organic compound in Kjeldhal's method liberated ammonia, which nutralised 60 ml of 0.1 N H_(2)SO_(4) solution. The percentage of nitrogen in the compound is |
| Answer» <html><body><p>`1.68`<br/>`16.8`<br/>`33.6`<br/>`8.4`</p>Solution :% of `<a href="https://interviewquestions.tuteehub.com/tag/n-568463" style="font-weight:bold;" target="_blank" title="Click to know more about N">N</a>=("No. of m.eq. of " 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>)" liberated"xx1.4)/("<a href="https://interviewquestions.tuteehub.com/tag/weight-1451304" style="font-weight:bold;" target="_blank" title="Click to know more about WEIGHT">WEIGHT</a> of organic compound")`<br/>`(60xx0.1xx1.4)/(0.5)=16.8 %`</body></html> | |
| 51957. |
0.5 g of impure ammonium chloride was heated with caustic soda solution to evolve ammonia gas, the gas is absorbed in 150 mL of N//5 H_(2)SO_(4) solution. Excess sulphuric acid required 20 mL of 1N NaOH for complete neutralization. The percentage of NH_(3) in the ammonium chloride is : |
| Answer» <html><body><p>0.68<br/>0.34<br/>0.48<br/>0.17</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> | |
| 51958. |
0.5 g of bleaching powder was suspended in water an excess of KI added. On acidifying with dilute H_(2)SO_(4) iodine was liberated which required 50 mL of N//10 hypo solution. Calculate the percentage of available chlorine in bleaching powder. |
| 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> | |
| 51959. |
0.5 g of an organic compound cantaining N on Kjeldahlising required 29 ml of N/5 H_(2)SO_(4) for complete neutralization of NH3 . The percentage of N in the compound is |
| Answer» <html><body><p>34.3<br/>16.2<br/>21.6<br/>14.8</p>Solution :`% ofN_2 = (28xx0.129xx100)/(1000xx 0.5)=16.2`</body></html> | |
| 51960. |
0.5 g of an organic compound containing nitrogen on Kjeldahlising required 29 ml of N/5 H_(2)SO_(4) for complete neutralisation of ammonia. The percentage of nitrogen in the compound is |
| Answer» <html><body><p>34.34<br/>40<br/>21.64<br/>14.84</p>Solution :`%<a href="https://interviewquestions.tuteehub.com/tag/n-568463" style="font-weight:bold;" target="_blank" title="Click to know more about N">N</a> = (1.4 <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> N 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>)/(<a href="https://interviewquestions.tuteehub.com/tag/w-729065" style="font-weight:bold;" target="_blank" title="Click to know more about W">W</a>)`</body></html> | |
| 51961. |
0.5 g of a sample of bleaching powder was suspended in water and excess KI is added. On acidifying with dil. H_(2)SO_(4). I_(2) was liberated which required 50 mL of N//10 hypo (Na_(2)S_(2)O_(3). 5H_(2)O). Calculate the percentage of available Cl_(2) in bleaching powder. |
| 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> | |
| 51962. |
0.5 g o fan impure sample of oxalate was dissolved in water and the solution made 100 mL on titration 10 mLof this solution requared 15 mL of N//20 KMnO_(4) solution . Calculate the percentage of pure oxalate in the sample |
| Answer» <html><body><p></p>Solution :The reduction half reaction is `C_(2)O_(4)^(2-) rarr 2CO_(2) +2e^(-)` <br/> `<a href="https://interviewquestions.tuteehub.com/tag/therefore-706901" style="font-weight:bold;" target="_blank" title="Click to know more about THEREFORE">THEREFORE</a>` Eq we of `C_(2)O_(4)^(2-)=(2xx12+4xx16)/(2)=44` <br/> Let `N_(1)` be the normality of the oxalte solution applying normality equation we have <br/> `15xx1//20(KmnO_(4))=N_(1)xx10(C_(2) O_(4))^(2-) therefore N_(1)=(15)/(10xx20)=3/40` <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/> Amount of pure `C_(2)O_(4)^(2-)` present in 100 mL =`(33/1000)xx100=0.33 <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/> But amount of `C_(2)O_(4)^(2-)` present in impure sample =0.5 g (given ) <br/> <a href="https://interviewquestions.tuteehub.com/tag/percentage-13406" style="font-weight:bold;" target="_blank" title="Click to know more about PERCENTAGE">PERCENTAGE</a> of pure <a href="https://interviewquestions.tuteehub.com/tag/oxalate-1144431" style="font-weight:bold;" target="_blank" title="Click to know more about OXALATE">OXALATE</a> `=(0.33)/(0.50)xx100=66`</body></html> | |
| 51963. |
0.5 g mixture of oxalic acid (H_(2)C_(2)O_4)and some sodium oxalate (Na_(2)C_(2)O_4)with some impurities requires 40 ml of 0.1M NaOH for complete neutralization and 6ml of 0.2 M KMnO_4for complete oxidation. Calculate the % of Na_2C_(2)O_4in the mixture |
| Answer» <html><body><p> `90%`<br/>` 26.8%`<br/>`40%`<br/>`<a href="https://interviewquestions.tuteehub.com/tag/50-322056" style="font-weight:bold;" target="_blank" title="Click to know more about 50">50</a>%`</p>Solution : No. of milli equivalents of `H_(2)C_(2)O_4` = No .of milli equivalents of NaOH = 4 No.of milli <br/> equivalents of `H_(2)C_(4)O_(4)+Na_(20)C_(2)O_(4)`= No. of <br/> milli equivalents of `KMnO_(4)= <a href="https://interviewquestions.tuteehub.com/tag/6-327005" style="font-weight:bold;" target="_blank" title="Click to know more about 6">6</a> xx 0.2 xx 5 = 6` <br/>Milli <a href="https://interviewquestions.tuteehub.com/tag/equivalent-446407" style="font-weight:bold;" target="_blank" title="Click to know more about EQUIVALENT">EQUIVALENT</a> of `Na_(2)C_(2)O_(4)= 2` <br/>Weight of `Na_(2)C_(2)O_(4)=2xx10^(-3)xx67 =0.134` g <br/> % `Na_(2)C_2O_4=(0.134)/(0.5)xx100=26.8`</body></html> | |
| 51964. |
0.5 g bleaching powder was suspended in water and excess KI is added. On acidifying with dil. H_(2)SO_(4), I_(2) was liberated which required 50 mL of N//10 hypo (Na_(2)S_(2)O_(3). 5H_(2)O) in presence of starch. The reactions involved are : I. CaOCl_(2)+H_(2)SO_(4) to CaSO_(4) to CaSO_(4)+H_(2)O +Cl_(2) II. 2KI+Cl_(2) to 2KCl+I_(2) III. 2Na_(2)S_(2)O_(3)+I_(2) to Na_(2)S_(4)O_(6)+2NaI In the reaction (I), which one is reduced ? |
| Answer» <html><body><p>`CaOCl_(<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/>`H_(2)SO_(<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/>Both<br/>none of these</p>Solution :N//A</body></html> | |
| 51965. |
0.5 g bleaching powder was suspended in water and excess KI is added. On acidifying with dil. H_(2)SO_(4), I_(2) was liberated which required 50 mL of N//10 hypo (Na_(2)S_(2)O_(3). 5H_(2)O) in presence of starch. The reactions involved are : I. CaOCl_(2)+H_(2)SO_(4) to CaSO_(4) to CaSO_(4)+H_(2)O +Cl_(2) II. 2KI+Cl_(2) to 2KCl+I_(2) III. 2Na_(2)S_(2)O_(3)+I_(2) to Na_(2)S_(4)O_(6)+2NaI In reaction (II), Cl_(2) acts as : |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/reducing-2981618" style="font-weight:bold;" target="_blank" title="Click to know more about REDUCING">REDUCING</a> <a href="https://interviewquestions.tuteehub.com/tag/agent-369049" style="font-weight:bold;" target="_blank" title="Click to know more about AGENT">AGENT</a><br/><a href="https://interviewquestions.tuteehub.com/tag/oxidising-2209005" style="font-weight:bold;" target="_blank" title="Click to know more about OXIDISING">OXIDISING</a> agent<br/>indicator<br/>both oxidising agent and indicator</p>Solution :N//A</body></html> | |
| 51966. |
0.5 g bleaching powder was suspended in water and excess KI is added. On acidifying with dil. H_(2)SO_(4), I_(2) was liberated which required 50 mL of N//10 hypo (Na_(2)S_(2)O_(3). 5H_(2)O) in presence of starch. The reactions involved are : I. CaOCl_(2)+H_(2)SO_(4) to CaSO_(4) to CaSO_(4)+H_(2)O +Cl_(2) II. 2KI+Cl_(2) to 2KCl+I_(2) III. 2Na_(2)S_(2)O_(3)+I_(2) to Na_(2)S_(4)O_(6)+2NaI Starch forms iodo-starch complex in the given titration. The colour of the complex will be : |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/green-476835" style="font-weight:bold;" target="_blank" title="Click to know more about GREEN">GREEN</a> <br/>blue<br/>pale yellow<br/>milky white</p>Solution :N//A</body></html> | |
| 51967. |
0.5 g bleaching powder was suspended in water and excess KI is added. On acidifying with dil. H_(2)SO_(4), I_(2) was liberated which required 50 mL of N//10 hypo (Na_(2)S_(2)O_(3). 5H_(2)O) in presence of starch. The reactions involved are : I. CaOCl_(2)+H_(2)SO_(4) to CaSO_(4) to CaSO_(4)+H_(2)O +Cl_(2) II. 2KI+Cl_(2) to 2KCl+I_(2) III. 2Na_(2)S_(2)O_(3)+I_(2) to Na_(2)S_(4)O_(6)+2NaI Percentage of available chlorine in bleaching powder is : |
| Answer» <html><body><p>`35.5%`<br/>`<a href="https://interviewquestions.tuteehub.com/tag/71-334184" style="font-weight:bold;" target="_blank" title="Click to know more about 71">71</a>%`<br/>`17.25%`<br/>`<a href="https://interviewquestions.tuteehub.com/tag/50-322056" style="font-weight:bold;" target="_blank" title="Click to know more about 50">50</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> :N//A</body></html> | |
| 51968. |
0.5 g bleaching powder was suspended in water and excess KI is added. On acidifying with dil. H_(2)SO_(4), I_(2) was liberated which required 50 mL of N//10 hypo (Na_(2)S_(2)O_(3). 5H_(2)O) in presence of starch. The reactions involved are : I. CaOCl_(2)+H_(2)SO_(4) to CaSO_(4) to CaSO_(4)+H_(2)O +Cl_(2) II. 2KI+Cl_(2) to 2KCl+I_(2) III. 2Na_(2)S_(2)O_(3)+I_(2) to Na_(2)S_(4)O_(6)+2NaI In the given titration, starch acts as : |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/oxidising-2209005" style="font-weight:bold;" target="_blank" title="Click to know more about OXIDISING">OXIDISING</a> agent<br/>indicator<br/>reducing agent<br/>catalyst</p>Solution :N//A</body></html> | |
| 51969. |
0.482 mol N_2 and 0.933 mol O_2 are placed in a 10 L vessel and allowed to form N_2O at constant temperature. Numerical valueof K_c for the reaction 2N_2(g)+O_2(g) leftrightarrow 2N_2O(g) is 2 times 10^-37 K mol^-1. Calculate the equilibrium concentration of nirous oxide. |
| 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> :`6.6 <a href="https://interviewquestions.tuteehub.com/tag/times-1420471" style="font-weight:bold;" target="_blank" title="Click to know more about TIMES">TIMES</a> 10^-21 <a href="https://interviewquestions.tuteehub.com/tag/mol-1100196" style="font-weight:bold;" target="_blank" title="Click to know more about MOL">MOL</a> L^-1`</body></html> | |
| 51970. |
0.48 g of a sample of a compound containing boron and oxygen contains 0.192 g of boron and 0.288 g of oxygen. What is the percentage composition of the compound ? |
| 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> :<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 compound <a href="https://interviewquestions.tuteehub.com/tag/taken-659096" style="font-weight:bold;" target="_blank" title="Click to know more about TAKEN">TAKEN</a> = 0.48 <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 boron present = 0.192 g <br/> Mass of oxygen present = 0.288 g <br/> Percent of B `= (("0.192 g"))/(("0.48g"))xx100=40%` <br/> Percentage of O `=(("0.288 g"))/(("0.48 g"))xx100=60%`.</body></html> | |
| 51971. |
0.46g of an organic compound was analysed. The increase in mass of CaCl_(2) U-tube was 0.54g and potash bulb was 0.88g. The percentage composition of the compound is |
| Answer» <html><body><p>`C=52.17%,H=13.04%,O=34.79%`<br/>`C=50%,H=50%`<br/>`C=32.19%,H=<a href="https://interviewquestions.tuteehub.com/tag/18-278913" style="font-weight:bold;" target="_blank" title="Click to know more about 18">18</a>.01%,O=49.8%`<br/>`C=72%,H=28%`</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 `H_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)O=0.54g` <br/> Mass of `CO_(2)=0.88g` <br/> % of `C=(12)/(<a href="https://interviewquestions.tuteehub.com/tag/44-316683" style="font-weight:bold;" target="_blank" title="Click to know more about 44">44</a>)xx(0.88)/(0.46)xx100=52.17%` <br/> % of `H=(2)/(18)xx(0.54)/(0.46)xx100=13.04%` <br/> % of `O=100-(52.17+13.04)=34.79`%</body></html> | |
| 51972. |
0.465 g of an organic substance gave on combustion 1.32 g of CO_(2) and 0.315g of H_(2)O. Calculate the percentage of carbon and hydrogen in the compound. |
| 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/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a> = 77.42%, <a href="https://interviewquestions.tuteehub.com/tag/h-1014193" style="font-weight:bold;" target="_blank" title="Click to know more about H">H</a> = 7.53 %</body></html> | |
| 51973. |
0.45g organic compound by combustion, give 0.792g CO_(2) and 0.324g water. For this compound ammonia obtain from 0.24g compound in Kjeldahl's method absorb in 50 ml, 0.25N, H_(2)SO_(4). In neutralisation 77.0 mL, 0.25N is used. Determine the empirical formula of compound NaOH |
| 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> :`C_(<a href="https://interviewquestions.tuteehub.com/tag/6-327005" style="font-weight:bold;" target="_blank" title="Click to know more about 6">6</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>)NO`</body></html> | |
| 51974. |
0.456 gof a metal gives 0.606gof its chloride. Calculate the equivalent mass of the metal. |
| Answer» <html><body><p></p>Solution :Mass of the metal `=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>)=0.606g` <br/> `<a href="https://interviewquestions.tuteehub.com/tag/therefore-706901" style="font-weight:bold;" target="_blank" title="Click to know more about THEREFORE">THEREFORE</a>` Mass of <a href="https://interviewquestions.tuteehub.com/tag/chlorine-915874" style="font-weight:bold;" target="_blank" title="Click to know more about CHLORINE">CHLORINE</a> `=W_(2)=0.606-0.456=0.15g` <br/> `0.15g ` of chlorine combine with `0.456 g` of metal <br/> `therefore 35.46 g` of chlorine will combine with <br/> `(0.456)/(0.15)xx35.46=107.76g "eq"^(-1)` <br/> Mass of chloride `=0.606-0.456=0.146g` <br/> `0.146 g` of chlorine combines with `0.456 g` of metal. <br/> `therefore35 .5 g` of chlorine will combines with `=(35.5xx0.456)/(0.146)` <br/> `=110.8g` of metal <br/> `therefore` equivalent mass of metal `=110 .<a href="https://interviewquestions.tuteehub.com/tag/8-336412" style="font-weight:bold;" target="_blank" title="Click to know more about 8">8</a> g "<a href="https://interviewquestions.tuteehub.com/tag/equ-446399" style="font-weight:bold;" target="_blank" title="Click to know more about EQU">EQU</a>"^(-1)` .</body></html> | |
| 51975. |
0.456 g of a metal gives 0.606 g of its chloride. Calculate the equivalent mass of the metal. |
| Answer» <html><body><p></p>Solution :Mass of the metal = `W_i = 0.456 <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 the metal chloride =` W_2 =0.606 g` <br/>`therefore ` Mass of chloride `= W_2 - W_1` <br/>`=0.606 - 0.456 = 0.15 g` <br/> 0.15 g of chloride <a href="https://interviewquestions.tuteehub.com/tag/combine-409630" style="font-weight:bold;" target="_blank" title="Click to know more about COMBINE">COMBINE</a> with 0.456g of metal.<br/>`therefore ` 35.46 g of chloride will combine with `(0.456)/(0.15) xx 35.46= 107.79g eq^(-<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> | |
| 51976. |
0.45 N & 0.6 N NaOH solution are mixed in 2:1 by volume. The amount of solute present in 1 lit of this solution is |
| Answer» <html><body><p>0.5 gm<br/>25 gm<br/>20 gm <br/>5 g</p>Solution :`(N_(1)V_(1)+N_(2)V_(2))/(V_(1)+V_(2))=<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/> `N=(0.45xx2+0.6xx1)/(3)=0.5N` <br/> implies 0.5 <a href="https://interviewquestions.tuteehub.com/tag/moles-1100459" style="font-weight:bold;" target="_blank" title="Click to know more about MOLES">MOLES</a> in 1 lit. = <a href="https://interviewquestions.tuteehub.com/tag/20g-293021" style="font-weight:bold;" target="_blank" title="Click to know more about 20G">20G</a></body></html> | |
| 51977. |
0.45 g of an organic compound when analysed by combustion gave 1.10 g carbon dioxide and 0.30 g water. Calculate the persentage of carbon and hydrogen in it. |
| 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> :Mass of the organic compound =0.45 g <br/> Mass of `CO_(2)` formed = 1.10 g <br/> Mass of `H_(2)O` formed = 0.30 g <br/> Percentage of `C=12/44 xx("Mass of carbon dioxide formed")/("Mass of compound")xx100` <br/> `=12/44xx((1.10 g))/((0.45 g))xx100=66.67 %` <br/> Percentage of `<a href="https://interviewquestions.tuteehub.com/tag/h-1014193" style="font-weight:bold;" target="_blank" title="Click to know more about H">H</a>=2/18 xx ("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> formed")/("Mass of compound")xx100` <br/> `=2/18xx((0.30 g))/((0.45 g))xx100=7.41 %`.</body></html> | |
| 51978. |
0.452 g of a metal nitrate gave 0.4378 g of its metal sulphate. Calculate the equivalent weight of the metal. |
| 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> :Mass of the metal nitrate taken = 0.452 g<br/> Mass of the metal sulphate obtained = 0.4378 g <br/> <a href="https://interviewquestions.tuteehub.com/tag/suppose-656311" style="font-weight:bold;" target="_blank" title="Click to know more about SUPPOSE">SUPPOSE</a>, equivalent mass of metal = E Using the relation,<br/> `("Mass of metal nitrate")/("Mass of metal sulphate") = ("Equivalent <a href="https://interviewquestions.tuteehub.com/tag/weight-1451304" style="font-weight:bold;" target="_blank" title="Click to know more about WEIGHT">WEIGHT</a> of metal nitrate")/("Equivalent weight of metal sulphate")` <br/> `0.452/0.4378 = (E+ 62)/(E + 48)` <br/> [Eq. weight of `NO_(3)^(-) = (14 + (<a href="https://interviewquestions.tuteehub.com/tag/16-276476" style="font-weight:bold;" target="_blank" title="Click to know more about 16">16</a> xx 3))/1 = 62` <br/> [Eq. weight of `SO_(4)^(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>-) = (32 + (16 xx 4))/2 = 48` <br/> or `0.452-0.4378 xx E = (0.4378 xx 62) -(0.452 xx 48) = 5.4476` <br/> or `E=5.4476/(0.452-0.4378) = 5.4476/0.0142 = 383.6`</body></html> | |
| 51979. |
0.45 g of an organic compound gave 0.44 g of CO_(2) and 0.09 g of H_(2)O. The molecular mass of the compound is 90 amu. Calculate the molecular formula. |
| Answer» <html><body><p><br/></p>Solution :Step I. Percentage of elements <br/> Percentage of carbon `= (12)/(44) <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> ("Mass of "CO_(2))/("Mass of compound")xx100` <br/> `=(12)/(44)xx(0.44)/(0.45)xx100=26.67` <br/> Percentage of <a href="https://interviewquestions.tuteehub.com/tag/hydrogen-22331" style="font-weight:bold;" target="_blank" title="Click to know more about HYDROGEN">HYDROGEN</a> `=(2)/(18)xx("Mass of" H_(2)O)/("Mass of compound")xx100` <br/> `=(2)/(18)xx(0.09)/(0.45)xx100=2.22` <br/>Percentage of oxygeb `= 100 - (26.67 + 2.22) = 100-28.89 = 71.11` <br/> Step II. Empirical formula of the compound <br/> `{:("<a href="https://interviewquestions.tuteehub.com/tag/element-969236" style="font-weight:bold;" target="_blank" title="Click to know more about ELEMENT">ELEMENT</a>","Percentage","Atomic mass","Gram atoms (Moles)","Atomic ratio (Molar ratio)","Simplest whole no. ratio"),("C",26.67,12,(26.67)/(12)=2.22,(2.22)/(2.22)=1.0,1),("H",2.22,1,(2.22)/(1)=2.22,(2.22)/(2.22)=1.0,1),("O",71.11,16,(71.11)/(16)=4.44,(4.44)/(2.22)=2.0,2):}` <br/> Empirical formula of the compound `= CHO_(2)`<br/> Step <a href="https://interviewquestions.tuteehub.com/tag/iii-497983" style="font-weight:bold;" target="_blank" title="Click to know more about III">III</a>. Molecular formula of the compound <br/> Empirical formula mass `= 12+1 +2 xx 16 = 45 u` <br/> Molecular mass = 90 <a href="https://interviewquestions.tuteehub.com/tag/amu-25369" style="font-weight:bold;" target="_blank" title="Click to know more about AMU">AMU</a> (Given) <br/> `n=("Molecular mass")/("Empirical formula mass")=((90u))/((45u))=2` <br/>Molecular formula `= n xx` Empirical formula `= 2 xx CHO_(2)=C_(2)H_(2)O_(4)`.</body></html> | |
| 51980. |
0.4422g of an organic compound was Kjeldahlised and ammonia evolved was absorbed in 50 mL of semi-molar (0.5 M) H_(2)SO_(4). The residual acid required 131 mL of 0.25 M NaOH. Determine the percentange of nitrogen in the compound. |
| 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> :`54.61%`</body></html> | |
| 51981. |
0.4037 g of an organic substance containing sulphur was heated with conc. nitric acid in a carius tube. On precipitation with BaCl_(2), 0.1963 g of BaSO_(4) was produced. Determine the percentage of sulphur in the compound. |
| 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> :`S = 7.66%`</body></html> | |
| 51982. |
0.40 g of an iodo-substituted organic compound gave 0.235 g of AgI by carius method. Calculate the percentage of iodine in the compound. (Ag = 108, I = 127). |
| 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> :Weight of <a href="https://interviewquestions.tuteehub.com/tag/organic-1138713" style="font-weight:bold;" target="_blank" title="Click to know more about ORGANIC">ORGANIC</a> substance (w) = 0.40 g <br/> Weight of silver iodine (x) = 0.<a href="https://interviewquestions.tuteehub.com/tag/235-1826120" style="font-weight:bold;" target="_blank" title="Click to know more about 235">235</a> g <br/> `"Percentage of iodine"=(<a href="https://interviewquestions.tuteehub.com/tag/127-271424" style="font-weight:bold;" target="_blank" title="Click to know more about 127">127</a>)/(235) xx (x)/(w) xx 100=(127)/(235) xx (0.235)/(0.40) xx 100=31.75%`</body></html> | |
| 51983. |
0.40 g of an organic compound containing phosphorus gave 0.555 g of Mg_(2)P_(2)O_(7) by usual analysis. Calculate the percentange of phosphorusin the organiccompound. |
| 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> :`38.75%`</body></html> | |
| 51984. |
0.4 mole of orthophosphoric acid and 1.0 mole of calcium hydroxide were allowed to react. Calculate the maximum number of moles of calcium phosphate formed. |
| Answer» <html><body><p></p>Solution :The stoichiometric equation of <br/> `3Ca(<a href="https://interviewquestions.tuteehub.com/tag/oh-585115" style="font-weight:bold;" target="_blank" title="Click to know more about OH">OH</a>)_(2)+2H_(3)PO_(4) to Ca_(3)(PO_(4))_(2)+3H_(2)O` <br/> 3 <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/ca-375" style="font-weight:bold;" target="_blank" title="Click to know more about CA">CA</a>(OH)_(2)="2 moles of "H_(3)PO_(4)` <br/> 1 mole of `Ca(OH)_(2)="0.67 mole of "H_(3)PO_(4)` <br/> 0.6 mole of `Ca(OH)_(2)="0.4 mole of "H_(3)PO_(4)` <br/> `H_(3)PO_(4)` is the limiting regent. Hence, salt formed is dependent on the availability of acid only. <br/> 2 moles of `H_(3)PO_(4)="1 mole of "Ca_(3)(PO_(4))_(2)` <br/> 0.4 moles of `H_(3)PO_(4)=?` <br/> Number of moles of calcium phosphate formed `=0.4 xx 1/2=0.2"mole"`</body></html> | |
| 51985. |
0.4 gm of polybasic acid H_(n)A (M.wt = 96) requires 0.5 gm NaOH for complete neutralisation. The number of replacable hydorgen atoms are (all the hydrogens are acidic) |
| 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> :m <a href="https://interviewquestions.tuteehub.com/tag/eq-446394" style="font-weight:bold;" target="_blank" title="Click to know more about EQ">EQ</a> <a href="https://interviewquestions.tuteehub.com/tag/acid-847491" style="font-weight:bold;" target="_blank" title="Click to know more about ACID">ACID</a> = M eq <a href="https://interviewquestions.tuteehub.com/tag/base-892693" style="font-weight:bold;" target="_blank" title="Click to know more about BASE">BASE</a> <br/> `(0.4)/(Nw//n)xx1000=(0.5)/(40//1)xx1000impliesn=3`</body></html> | |
| 51986. |
0.4 g of an organic compound in Dumas method, gave 50 ml of nitrogen collected at 300 K and 745 mn pressure Calculate the percentage composition of nitrogen in the compound. Aqueous tension at 300 K is 15 mm. |
| Answer» <html><body><p>0.4332<br/>0.6439<br/>0.2242<br/>0.1367</p>Solution :`V_0 =(739xx50 xx273)/(<a href="https://interviewquestions.tuteehub.com/tag/300-305868" style="font-weight:bold;" target="_blank" title="Click to know more about 300">300</a> <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/760-335611" style="font-weight:bold;" target="_blank" title="Click to know more about 760">760</a>)=43.7 " <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/> `" % of " N_2=(28xx43.7 <a href="https://interviewquestions.tuteehub.com/tag/xx100-3292680" style="font-weight:bold;" target="_blank" title="Click to know more about XX100">XX100</a>)/(22,400 xx 0.4) =13.67`</body></html> | |
| 51987. |
0.4g of a compound on complete combustion gave 56ml of CO_(2) at 760mm and 0^(@)C. The percentage of carbon in the compound is |
| Answer» <html><body><p>50<br/>60<br/>27.5<br/>7.5</p>Answer :D</body></html> | |
| 51988. |
0.4 g of an organic compound gave 0.188g of silver bromide by a halogen estimation method. The percentage of bromine in the compound is (at . Wts of Ag= 108, Br = 80) |
| Answer» <html><body><p>20<br/>40<br/>46<br/>60</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> | |
| 51989. |
0.4g ofan organiccompoundgave0.188g ofsilverbromidebyhalogenestimationmaethodThepercentageof brominein thecompoundis ___.( At .Massof Ag =108 , Br=80) |
| Answer» <html><body><p>0.398<br/>0.46<br/>0.2<br/>0.4</p>Solution :`0.<a href="https://interviewquestions.tuteehub.com/tag/188-280322" style="font-weight:bold;" target="_blank" title="Click to know more about 188">188</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> of <a href="https://interviewquestions.tuteehub.com/tag/silver-644119" style="font-weight:bold;" target="_blank" title="Click to know more about SILVER">SILVER</a>" -= (0.188xx80)/(188) =0.08g " of "Bra_2` <br/> `therefore% of <a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a> r_2 =(0.08)/(0.4) xx 100=20`</body></html> | |
| 51990. |
0.3g of an organic compound on combustion liberated 0.18g of water vapour and 0.44 g of carbonioxide. Calculate the percentage composition of the compound |
| Answer» <html><body><p></p>Solution :Weight of the compound =W = 0.3g <br/> Weight of `CO_(2)= 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>) = 0.44g` <br/> Weight of `H_(2)O = W_(2) = 0.18g` <br/> % of carbon `=(W_(1) xx 12 xx 100)/(W xx 44) = (0.44 xx 12 xx 100)/(0.3 xx 44) = 40.0%` <br/> % of hydrogen `= (W_(2) xx 2xx 100)/(W xx <a href="https://interviewquestions.tuteehub.com/tag/18-278913" style="font-weight:bold;" target="_blank" title="Click to know more about 18">18</a>) = (0.18 xx 2 xx 100)/(0.3 xx 18) = 6.67%` <br/> After calculating the percentage <a href="https://interviewquestions.tuteehub.com/tag/composition-22493" style="font-weight:bold;" target="_blank" title="Click to know more about COMPOSITION">COMPOSITION</a> of the given elements in the compound, if their sum is not 100, then the difference is assumed as oxygen <br/> `:.` % of oxygen `=[100-(40+6.67)]= 53.33%`</body></html> | |
| 51991. |
0.395 g of an organic compound by carius method for the estimation of sulphur gave 0.582 g of BaSO_(4). Calculate the percentage of sulphur in the compound. |
| 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> :Percentage of sulphur `=32/233xx("<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 "BaSO_(<a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>))/("Mass of <a href="https://interviewquestions.tuteehub.com/tag/compound-926863" style="font-weight:bold;" target="_blank" title="Click to know more about COMPOUND">COMPOUND</a>")xx100=32/233xx0.582/0.395xx100=20.24 %`</body></html> | |
| 51992. |
0.37g of an alcohol (A) when treated with CH_(3)MgI liberates 112 ml of CH_(4) at STP. Alcohol (A) on dehydration gives alkene (B) which on ozonolysis gives acetone as one of the product along with (C). (A) when treated with benzene in acidic medium given (D) which does not give benzoc acid on oxidation. (A) on oxidation gives acid (E) having same number of C-atoms. Q. Structure of B is |
| Answer» <html><body><p>`CH_(3)-CH_(<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/ch-913588" style="font-weight:bold;" target="_blank" title="Click to know more about CH">CH</a>=CH-CH_(3)`<br/><img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/NAR_CHM_XI_V06_C02_E01_160_O02.png" width="30%"/><br/>`CH_(3)-overset(CH_(3))overset(|)(CH)-CH_(2)<a href="https://interviewquestions.tuteehub.com/tag/oh-585115" style="font-weight:bold;" target="_blank" title="Click to know more about OH">OH</a>`<br/>`CH_(3)-CH_(2)-CH=CH_(2)`</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> | |
| 51993. |
0.37g of an alcohol (A) when treated with CH_(3)MgI liberates 112 ml of CH_(4) at STP. Alcohol (A) on dehydration gives alkene (B) which on ozonolysis gives acetone as one of the product along with (C). (A) when treated with benzene in acidic medium given (D) which does not give benzoc acid on oxidation. (A) on oxidation gives acid (E) having same number of C-atoms. Q. Structure of A is |
| Answer» <html><body><p>`CH_(3)-CH_(2)-CH-CH_(2)-OH`<br/>`CH_(3)-overset(CH_(3))overset(|)(C)H-CH_(2)OH`<br/>`CH_(3)-CH_(2)-<a href="https://interviewquestions.tuteehub.com/tag/underset-3243992" style="font-weight:bold;" target="_blank" title="Click to know more about UNDERSET">UNDERSET</a>(OH)underset(|)(CH)-CH_(3)`<br/>`CH_(3)-underset(OH)underset(|)overset(CH_(3))overset(|)(C)-CH_(3)`</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> | |
| 51994. |
0.37g of an alcohol (A) when treated with CH_(3)MgI liberates 112 ml of CH_(4) at STP. Alcohol (A) on dehydration gives alkene (B) which on ozonolysis gives acetone as one of the product along with (C). (A) when treated with benzene in acidic medium given (D) which does not give benzoc acid on oxidation. (A) on oxidation gives acid (E) having same number of C-atoms. Q. From given data molecular formula of (A) is |
| Answer» <html><body><p>`C_(5)H_(11)<a href="https://interviewquestions.tuteehub.com/tag/oh-585115" style="font-weight:bold;" target="_blank" title="Click to know more about OH">OH</a>`<br/>`C_(<a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>)H_(<a href="https://interviewquestions.tuteehub.com/tag/9-340408" style="font-weight:bold;" target="_blank" title="Click to know more about 9">9</a>)OH`<br/>`C_(<a href="https://interviewquestions.tuteehub.com/tag/6-327005" style="font-weight:bold;" target="_blank" title="Click to know more about 6">6</a>)H_(<a href="https://interviewquestions.tuteehub.com/tag/13-271882" style="font-weight:bold;" target="_blank" title="Click to know more about 13">13</a>)OH`<br/>`C_(4)H_(7)OH`</p>Answer :B</body></html> | |
| 51995. |
0.3780g of an organic chloro compound gave 0.5740g of silver chloride in Carius estimation. Calculate the percentage of chlorine present in the compound (Ag= 108, Cl= 35.5) |
| 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> :`underset(35.5g)(Cl) overset(+)rarr underset(143.5g)(AgCl)` <br/> molecular mass of AgCl= 108 + 35.5 = 143.5g <br/> 143.5g AgCl contain 35.5g <a href="https://interviewquestions.tuteehub.com/tag/chlorine-915874" style="font-weight:bold;" target="_blank" title="Click to know more about CHLORINE">CHLORINE</a>. <br/> `therefore` 0.5740g AgCl contain, <br/> Chlorine `=(35.5 xx 0.5740)/(143.5)gm` <br/> % `Cl = (35.5)/(143.5) xx (0.5740)/(0.3780) xx <a href="https://interviewquestions.tuteehub.com/tag/100-263808" style="font-weight:bold;" target="_blank" title="Click to know more about 100">100</a>` <br/> =37.5661= 37.57</body></html> | |
| 51996. |
0.3780 g of an organic compound gave 0.5740 g of silver chloride in Carius estimation. Calculate the percentage of chlorine in the compound. |
| 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> :Mass of the <a href="https://interviewquestions.tuteehub.com/tag/compound-926863" style="font-weight:bold;" target="_blank" title="Click to know more about COMPOUND">COMPOUND</a> `= 0.3780 g` <br/> Mass of <a href="https://interviewquestions.tuteehub.com/tag/silver-644119" style="font-weight:bold;" target="_blank" title="Click to know more about SILVER">SILVER</a> chloride `=0.5740 g` <br/> <a href="https://interviewquestions.tuteehub.com/tag/percentage-13406" style="font-weight:bold;" target="_blank" title="Click to know more about PERCENTAGE">PERCENTAGE</a> of <a href="https://interviewquestions.tuteehub.com/tag/chlorine-915874" style="font-weight:bold;" target="_blank" title="Click to know more about CHLORINE">CHLORINE</a> `=35.5/143.5xx("Mass of silver chloride")/("Mass of compound")xx100` <br/> `=35.5/143.5xx((0.5740 g))/((0.3780 g))xx100=37.57 %`</body></html> | |
| 51997. |
0.378 g of an organic acid gave on combustion 0.264 g of carbon dioxide and 0.162 g of water vapour . Calculate the percentage of C and H. |
| 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 <a href="https://interviewquestions.tuteehub.com/tag/organic-1138713" style="font-weight:bold;" target="_blank" title="Click to know more about ORGANIC">ORGANIC</a> compound =0.378 g <br/> Mass of `CO_(2)` formedc =0.264 g <br/> Mass of `H_(2)O` formed=0.162 g <br/> (i) <a href="https://interviewquestions.tuteehub.com/tag/percentage-13406" style="font-weight:bold;" target="_blank" title="Click to know more about PERCENTAGE">PERCENTAGE</a> of carbon <br/> <a href="https://interviewquestions.tuteehub.com/tag/44-316683" style="font-weight:bold;" target="_blank" title="Click to know more about 44">44</a> g of `CO_(2)` contains carbon =12 g <br/> 0.264 g of `CO_(2)` contains carbon =`12/44 xx 0.264 =0.072 g ` <br/> Percentage of hydrogen <br/> 18 g of `H_(2)O` contains hydrogen = 2g <br/> 0.16.2 of `H_(2)O` contains hydrogen =`2/18xx0.162 =0.018 g ` <br/> Percentage of hydrogen =`0.018/0.378 xx100 =4.76 % `</body></html> | |
| 51998. |
0.3780 g of an organic chloro compound gave 0.5740 g silver chloride in Carius estimation. Calculate the percentage of chlorine present in the compound. |
| Answer» <html><body><p></p>Solution :Here, the <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 the substance taken = 0.3780 g <br/> Mass of <a href="https://interviewquestions.tuteehub.com/tag/agcl-368919" style="font-weight:bold;" target="_blank" title="Click to know more about AGCL">AGCL</a> <a href="https://interviewquestions.tuteehub.com/tag/formed-464209" style="font-weight:bold;" target="_blank" title="Click to know more about FORMED">FORMED</a> = 0.5740 g <br/>Now <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> of `AgCl -= 1` g atom of Cl or (108 + 35.5) = 143.5 g of `AgCl -= 35.5 g` of Cl <br/> Applying the relation, Percentage of chlorine `= (35.5)/(143.5) xx ("Mass of AgCl formed")/("Mass of substance taken") xx 100` <br/> `= (35.5)/(143.5) xx (0.5740)/(0.3780) xx 100 = 37.566%`.</body></html> | |
| 51999. |
0.3780 g an organic compound gave 0.5740 g AgCl in Carius method. Calculate the percentage of chlorine in it. |
| 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> :`(35.5)/143.5 <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> (0.5740)/0.3780 xx <a href="https://interviewquestions.tuteehub.com/tag/100-263808" style="font-weight:bold;" target="_blank" title="Click to know more about 100">100</a> = 37.57% <a href="https://interviewquestions.tuteehub.com/tag/chlorine-915874" style="font-weight:bold;" target="_blank" title="Click to know more about CHLORINE">CHLORINE</a>'</body></html> | |
| 52000. |
0.378 g od an organic compound containing carbon and hydrogen was subjected to cobustion by Liebig's method , the CO_(2) and H_(2)Oformed were passed through potash bulbs and CaCl_(2) (anhydrous) tube. At end of the experiment, the increase in the respective weights were 0.264 g and 0.162 g. Calculate the percentage of carbon and hydrogen. |
| Answer» <html><body><p><br/></p>Solution :Increase in weight of <a href="https://interviewquestions.tuteehub.com/tag/potash-1161117" style="font-weight:bold;" target="_blank" title="Click to know more about POTASH">POTASH</a> <a href="https://interviewquestions.tuteehub.com/tag/bulbs-905371" style="font-weight:bold;" target="_blank" title="Click to know more about BULBS">BULBS</a> (mass pf `CO_(2)` evolved) `=0.264 <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/> Increase in weight of `CaCl_(2)` tube (mass of `H_(2)O` forward) `=0.162 g` <br/> `%` of `C=12/44xx("Mass of "CO_(2))/("Mass of compound")xx100=12/44xx0.264/0.378xx100= 19.05%` <br/> `%` of `H=12/18xx("Mass of "H_(2)O)/("Mass of compound")xx100=2/18xx0.162/0.378xx100= 4.76 %`</body></html> | |