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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
0.984 gm of the chloroplatinate of a diacid base gave 0.39 gm of platinum. Calculate the molecular mass of the base. |
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Answer» Mass of platinichloride taken `=0.984gm` Mass of platinum left `=0.39gm` Step 1. To calculate the equivalent mass of the base. Let the equivalent mass of the base be B. Molecular mass of the platininchloride `(B_2H_2PtCL_6)` `=2B+410`. Now, `("molecular mass of chloroplatinichloride")/("atomic mass of platinum")` `=("mass of platinichloride taken")/("mass of platinum left")` or `(2B+410)/(195)=(0.984)/(0.39)` or `B=(1)/(2)((0.984)/(0.39)xx195-410)=41` Thus, the equivalent mass of the base be `41` Step 2: to calculate the molecular mass of the base. Acidity of the base `=2` Molecular mass of base `=`equivalent mass of base`xx`acidity `=41xx2=82` Thus, the molecular mass of base is `82`. |
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| 2. |
Will `CCl_4` give white precipitate of `AgCl` on heating with nitrate? Give reason for your answer |
| Answer» `CCl_4` will not give a white precipitate of `AgCl` when it is treated with a solution of silver nitrate. It is because `CCl_4` is a covalent compound and hence does not ionise to give `Cl` ions, which is the essential requirements for the `AgNO_3` test. | |
| 3. |
An organic compound contains `4%` sulphur. Its minimum molecular weight is:A. 200B. 400C. 800D. 1600 |
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Answer» Correct Answer - C Four grams S is in 100 gm of compound. 32 gm S is in `(100xx32)/(4)=800gm`. |
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| 4. |
Indulin contains `3.4%` suplhur.The minimum molecular mass of insulin is:A. 940B. 350C. 470D. 560 |
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Answer» Correct Answer - A 3.4 gm S in in 100 gm of compound 32 gm S is in `(100xx32)/(3.4)=941.7approx940` |
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| 5. |
The empirical formula of anorganic compound is `CH_2`.The mass of 1 mol of it is 42 gm.The molecular formula of the compound is:A. `C_4H_8`B. `C_2H_4`C. `C_3H_6`D. `CH_2` |
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Answer» Correct Answer - C E.F.W.`=CH_2=12+2=14` `n=(Mw)/(E.F.W.)=(42)/(14)=3` `MF=C_3H_6` |
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| 6. |
Liquid benzene `(C_6H_6)` burns in oxygen according to `2C_6H_6(1)+15O_2(g)rarr12CO_2(g)+6H_2O_(g)` How many litres of `O_2` at STP are needed for complete conbustion of39 gm of liquid benzene?A. 11.2 litresB. 74 litresC. 84 litresD. 22.4 litres |
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Answer» Correct Answer - C `2C_6H_6+15O_2rarr12CO_2+6H_2O` `2xx78gmimplies15xx22.4` litre of `O_2` at STP `39gmimplies(15xx22.4xx39)/(2.79)=84` litre |
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| 7. |
Which of the following sodium compound is/are formed when as organic compound containing both nitrogen and sulphur is fused with sodium?A. Suphite and cyanideB. ThiocyanateC. Cyanide and suphideD. nitrate and suplhide |
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Answer» Correct Answer - B `Na` reacts with C,N, andS to form `NaCNS` (sodium thiocynate). |
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| 8. |
Explain the rason for the fusion of an organic compound with metallic sodium for testing nitrogen, sulphur and halogens |
| Answer» The elements like `N,S,P` or halogens in the organic compounds are presents in the form of covalent bond and therefore they are converted into ionic species (e.g. `Coverset(o-)N,S^(2-),PO_4^(3-)`, and `X^(o-)` ions) by fusing with Na. these ionic species can be easily detected by ionic reaction. | |
| 9. |
Which of the following carbocation is most stable?A. `(CH_3)_3C.overset(o+)CH_2`B. `(CH_3)_3overset(o+)C`C. `CH_3CH_2overset(o+)CH_2`D. `CH_3overset(o+)CHCH_2CH_3` |
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Answer» Correct Answer - B `3^@` corbocation |
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| 10. |
One litre of a mixture of CO and `CO_2` is passed through red-hot charcoal. The volume now becomes 1.6 litre. Find the composition of the mixture by volume. |
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Answer» Let `CO_2=x`litre `CO=(1-x)` litre `underset(xlitre)CO_2overset(C)tounderset(2xlitre)2CO` `underset(x)CO_2+underset(1-x)Co overset(C)tounderset(2x)2CO+underset(1-x litre)(CO)` `2x+(1-x)=1.6` `x=0.6litre` volume of `CO_2=0.6` litre Volume of `CO=(1-0.6)=0.4litre` |
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| 11. |
Twenty millilitres of a gaseous hydrocarbon required 400 ml of iar for complete cumbustion. The air contains `20%` by explosion and cooling was found to be 380 ml. Q. volume of residual nitrogen is:A. 300 mlB. 310 mlC. 320 mlD. 330 ml |
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Answer» Correct Answer - C Volume of residual `N_2=(400-80)=320ml` |
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| 12. |
Fifty millilitres of pure and dry `O_2` was subjected to silent electric discharge and on cooling to the original temperature, the volume of the ozonised oxygen was found to be 47 ml. The gas was then absorbed in turpentine oil the volume of the remainig gas was found to e 41 ml. Find the molecular formula of ozon. |
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Answer» Volume of `O_2=50ml` Let the formula of ozone be `O_x` Volume of `O_2` (not converted to `O_x)+`ozone produced `=47ml` Volume of `O_2` left (not converted to `O_x)=41ml` Since ozone is absorbed in turpentine oil, Volume of `O_2` converted to `O_x=50-41` `=9ml` Volume of ozone produced `=47-41` `=6ml` `{:(O_2,rarr,O_x),(9ml,,6ml),(9 mol es,,6 mol es):}` Appying `POAC` rule (principle of atom conservation) for the O atom, we get Number of O atoms `xx` moles of `O_2=` Number of O atoms `xx` moles of `O_x` `=2xx9=x xx6` `impliesx=3` Hence, the formula of ozone is `O_3`. |
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| 13. |
fifty millilitre of a mixture of `CO` and `CH_4` was exploded with 85 ml of `O_2`. The volume of `CO_2` produced was 50 ml. Calculate the percentage composition of the gaseous mixture. |
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Answer» let the volume of `CO=xml` Volume of `CH_4=(50-x)ml` `underset(xml)CO+(1)/(2)O_2rarrunderset(xml)(CO_2)` .(i) `underset(50-x)CH_4+underset(2(50-x))+2O_2rarrunderset(50-x)CO_2+underset(-)2H_2O` .(ii) Total volume of `O_2=(x)/(2)+2(50-x)=85` `x=10ml` Volume of `CO=10ml` Volume of `CH_4=(50-10)=40ml` percentage of `CO=(10xx100)/(50)=20%` Percentage of `CH_4=(100-20)=80%` |
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| 14. |
An organic substance `(0.2115gm)` on complete combution gave 0.4655 gm of carbon dioxide and 0.2533 gm of water. Determine the percentage composition of the compound. |
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Answer» Mass of `CO_2` formed `=0.4655 gm` mass of `H_2O` formed `=0.2533gm` Mass of the organic substance taken `=0.2115gm` percentage of carbon `=(12)/(44)xx("mass of "CO_2)/("mass of compound")xx100` `=(12)/(44)xx(0.4655)/(0.2155)xx100=60.03` percentage of hydrogen `=(2)/(18)xx("mass of "H_2O)/("mass of compound")xx100` `=(2)/(18)xx(0.2533)/(0.2155)xx100=13.30` percentage of oxygen `=100-(60.3+13.30)=26.67` Composition of the substance : `C=60.03%,H=13.30%,` and `O=26.67%` |
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| 15. |
0.2475 gm of an organic substance gave on combustion 0.495 gm of `CO_2` and `0.2025gm` of `H_2O`. Calculate the percentange of carbon and hydrogen in it. |
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Answer» Here mass of the organic substance taken `=0.2475gm` mass of `CO_2` formed `=0.4950gm`, mass of `H_2O` formed `=0.2025gm`. (i) percentage of carbon in the compound `=(12)/(44)xx0.4950xx(100)/(0.2475)=54.54` (ii). Percentage hydrogen in the compound `=(2)/(18)xx0.2025xx(100)/(0.2475)=9.09` |
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| 16. |
0.5264 gm silver bromide is obtained from 0.5124 gm of an organic compound. Calculate the percentage of bromine in the compound. |
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Answer» Mass of organic compound `W=0.5124gm` Mass of silver bromide, `W_1=0.5264gm` `AgBr-=Br` percentage of bromine `=(80)/(188)xx(W_1)/(W)xx100` `=(80)/(188)xx(0.5264)/(0.5124)xx100=43.71` |
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| 17. |
A duma bulb full of air weighs 22.567 gm at `20^@C` and 755 mm pressure. Full of vapours of a substance at `120^@C` and the same pressure. It weighs 22.8617 gm. The capacity of the bulb is 200 ml. Find out the molecular mass of the substance. [density of air `=0.00129(gm)/(ml)`] |
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Answer» `{:("volume of bulb"=200ml,,V_2=?),(T_1=(20+273)=293K,,T_2=273K),(P_1=755mm,,P_2=760mm):}}` STP conditions so, `V_2=`volume of bulb at STP `=(200xx755)/(293)xx(273)/(760)=185.122ml` mass of air `=V_2xx0.00129=185.122xx0.00129` `=0.2388gm` mass of empty bulb `=(22.567-0.2388)` `=22.3282gm` mass of vapours `=(22.8617-22.3282)` `=0.5335gm` Let the volume of vapours at STP be V. `V=(200xx755)/(393)xx(273)/(760)=138ml` molecular mass of the substance `=("mass of vapours")/("volume of vapours at STP")xx22400` `=(0.5335)/(138)xx22400=86.69` |
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| 18. |
0.38 gm of a silver salt of a dibasic acid on igition gave 0.27 gm of silver. Calculate the molecular mass of the acid. |
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Answer» Mass of silver salt `=0.38gm` Mass of silver `=0.27gm` `("equivalent mass of silversalt")/("equialent mass of silver")=("mass of silver salt")/("mass of siver")` Equivalent mass of silver salt `=(0.38)/(0.27)xx108` `E_107=(0.38)/(027)xx108` (`E=` Equivalent mass of acid) `E=[(0.38)/(0.27)xx108-107]=45` Molecular mass of the acid `=` equivalet mass `xx` basicity `=45xx2=90` |
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| 19. |
0.5 gm of an organic substance containing prosphorous was heated with sonc. `HNO_3` is the carius tube. The phosphoric acid thus formed was preciopitated with magnesia mixture `(MgNH_4PO_4)` which on ignition gave a residue of 1.0 gm of magnesium phrrophosphate `(Mg_2P_2O_7)`. The precentage of phosphorous in the organic compound is:A. `55.85%`B. `29.72%`C. `19.18%`D. `20.5%` |
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Answer» Correct Answer - A `P+HNO_3toH_2PO_4underset((NH_4Cl+MgCl_2))toMgNH_4undersetunderset(Mg_2P_2O_7)(darrtriangle)(PO_4)` Mw of `Mg_2P_2O_7=24xx2+31xx2+16xx7=222` percentage of `P=(62)/(222)xx("weight of "Mg_2P_2O_7)/("weight of compound")xx100` `=(62)/(222)=(1.0)/(0.5)xx100=55.85%` |
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| 20. |
0.1693 gm of a volatile substance when vapourised displaced 58.9 cm of air measured at `27^@C` and 746 mm pressure. Calculate the molecular mass of the substance. (Aqueous tenstion at `27^@C=26.7mmHg`.) |
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Answer» To convert the volume at experimental conditions to volume at STP. `{:("Experimental condition", "At STP"),(P_1=746-26.7=719.3mm,P_2=760mm),(V_1=58.9ml,V_2=?),(T_1=273+27=300K,T_2=273K):}` Substituting these values in the gas equation, `(P_1V_1)/(T_1)=(P_2V_2)/(T_2)`, we get `(719.3mmxx58.9ml)/(300K)=(760mmxxV_2ml)/(273K)` `V_2=(719.3mmxx58.9mlxx273K)/(300Kxx760mm)=50.73ml` molecular mass `=("mass of substance"xx22400)/("volume of displaced air at "STP)` `=(0.1693xx22400)/(50.73)=74.75gm` |
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| 21. |
The best and latest technique for isolation, purification, and separtion of organic compounds isA. crystallisationB. distillationC. sublimationD. chromatography |
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Answer» Correct Answer - D Chromatography is the best and latest technique`. |
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| 22. |
(a) Two volatile counpunds differ in their boiling points by 20 K, how will they be separated? (b) What types of compounds are purified by sublimation? (c) How will `I_2` be separated from `KCl`? (d) How are `o-` and `p-` nitro phenols separated? (e) How is aniline purified? (f). How is a mixture of maphthalene and kerosene oil separated? |
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Answer» (a). Fractional distillation. (b). Substances whose vapour pressure become equal to the atmospheric pressure much below their boiling points. (c). Either by sublimation or by extraction with `C Cl_4` followed by evaporation (d) `o-` Nitrophenol is stream volatile while `p-` nitrophenol is not hence they are separated by steam distillation. (e). Vacuum distillation or steam distillation. (f). By simple distillation. |
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| 23. |
An organic compound contains `69%` carbon and `4.8%` hydrogen, the remainder being oxygen. Calculate the masses of carbon dioxide and water produced when 0.20 gm of this substance is subjected to complete combustion. |
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Answer» precentage of `C=(12)/(44)xx("mass of "CO_2" formed")/("mass of substance taken")xx100` `4.8=(2)/(18)xx("mass of "H_2O" formed")/("mass of substance taken")xx100` or mass of water found `=(4.8xx18xx0.2)/(2xx100)=0.0864gm` |
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| 24. |
Why is solution of potassium hydroxide used to absorb carbon dioxid evolved during the estimation of carbon present in an organic compound? |
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Answer» Carbon dioxide `(CO_2)` is slightly acidic in nature. Hence `KOH`, which is a strong alkali dissolves `CO_2` evolved from the compound containing carbon and reacts with it to form `K_2CO_3`. Increase in the weight of potash bulbs will give the weight of `CO_2` evloved. `2KOH(aq.)+CO_2(g)rarrK_2CO_3+H_2O(1)` |
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| 25. |
10 mL of a mixture of `CH_(4),C_(2)H_(4) and CO_(2)` was exploded with excess oxygen. After explosion, there ws a contraction of 17 mL on cooling and there was a further contraction of 14 mL on treatment with KOH. Find out the composition of the mixture. |
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Answer» Let the volume of `CH_4=xml` Volume of `(CH_2=CH_2)=yml` Volume of `CO_2=[10-(x+y)]ml` (i) `CH_4(g)+2O_2(g)rarrCO_2(g)+2H_2O(l)` (ii) `CH_2=CH_2(g)+3O_2(g)rarr2CO_2(g)+H_2O(l)` Volume of `CO_2` absorbed by `KOH=14ml` Volume of `CO_2=x+2y+[10-(x+y)]=14` `y=4ml` volume of gaseous reactant`=`Volume of gaseous products `+` contraction `x+2x+y+3y=x+2y+17` `3x+4y-x-2y=17` `2x+2y=17` `2x+2xx4=17` `2x=9` `x=4.5ml` Volume of `CH_4=4.5ml` volume of `(CH_2=CH_2)=4ml` volume of `CO_2=10-(4.5+4)` `=1.5ml` |
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| 26. |
1.575 gm of an organic acid was dissolved in 250 ml of water Further, 20 ml of this solution required 16 ml of `(N)/(8)` alkali solution for complete neutralisation. If the basicity of the acid is 2, find its molecular mass. |
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Answer» 16 ml `(N)/(8)` alkali solution `=20ml` of acid solution 2 ml 2 N alkali solution `=20ml` of acid solution `(2xx250)/(20)ml1N` alkali solution `=`250 ml of acid solution 25 ml 1 N alkali solution `=1.575 gm ` acid 1000 ml 1 N alkali solution `=(1.575)/(25)xx1000gm` acid equivalent mass fo the acid `=63gm acid` molecular mass of the acid `=63xx2=126` |
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| 27. |
A mixture of ethylene and excess of `H_2` has a pressure of 600 mm Hg. The mixture was passed over nickel catalyst to convert ethylene to ethane. The pressure of the resultant mixture at the similar condition of temperature and volume dropped to 400 mm Hg. the fraction of `C_2H_4` by volume dropped to 400 mm Hg. The fraction of `C_2H_4` by volume in the original mixture is:A. 1/3rd of the total volumeB. 1/4th of the total volumeC. 2/3rd of the total volumeD. 1/2 of the total volume |
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Answer» Correct Answer - A Let `n` mol of `(C_2H_4+H_2)` and `x` mol of `C_2H_4` `H_2=(n-x)mol` `underset(x)(C_2H_4)+underset(x)(H_2)rarrunderset(x mol)(C_2H_4)` After reaction `(C_2H_6+H_2` left) `x+n-x-x=n-x` [Total `H_2=(n-x),H_2` reacted `=x]` `H_2` left `=(n-x-x)` `n=600,n-x=400` `(n)/(n-x)=(600)/(400)` `x=(n)/(3)` volume of `C_2H_4` `=(1)/(3)` rd of total volume |
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| 28. |
Six hundred millitres of ozonised oxygen STP was found to weigh 1 gm. What is the volume of ozone in the ozonised oxygen?A. 200 mlB. 150 mlC. 100 mlD. 50 ml |
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Answer» Correct Answer - A Let the volume of `O_3` be `x` ml Volume of `O_2` present `=(600-x)ml` 22400ml of `O_3` and `O_2` at STP will weigh 48 and 32 gm respectively. The weight of `xml` of `O_3=(x xx48)/(22400)gm` The weight of `(600-x)ml` of `O_2=((600-x))/(22400)xx32` Total weight of ozonised `O_2(600ml)` is `(48x)/(22400)+((600-x)xx32)/(22400)=1.0` `x=200ml` |
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| 29. |
The weight of 1 litre of ozonised at STP was found to be 1.5 gm. When 100 ml of this mixture at STP was treated with turpentine oil the volume was reduced to 90 ml. The molecular weight of ozone isA. 49B. 47C. 46D. 47.9 |
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Answer» Correct Answer - C Volume of `O_3` in 100 ml of ozonised `O_2` `=100-90=10ml` (dissolved in turpentine) Volume of `O_3` in 1 litre of ozonised `O_2=(10xx100)/(100)` `=100ml` volume of `O_2` in 1 litre `=1000-100=900ml` Weight of `900ml` of `O_2` at S`TP=(900xx32)/(22400)` `=1.286gm` weight of `100 ml` of `O_3` at STP weighs`=0.214gm` `22400ml` of `O_3` at STP weghs`=(0.214xx22400)/(100)` `=47.94gm` Molecular weight of `O_3=47.94` |
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| 30. |
0.92 gm of an organic compound containing carbon, hydrogen, and oxygen was analysed by combustion method. The increase in the mass of the U-tube and the potash bulbs at the end of the operation was found to the 1.08 gm and 1.76 gm respectivley. Determine the percentage composition of the compound. |
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Answer» Increase in the mass of U-tube `=1.08gm` Mass of water formed `=1.08gm` since 18 gm of water contains 2 gm hydrogen 1.08 gm of water contains `(2)/(18)xx1.08gm` hydrogen mass of the compound taken for analysis `=0.92gm` Percentage of hydrogen `=(2)/(18)xx(1.08)/(0.92)xx100=13.04` Increase in the mass of potash bulbs `=1.76gm` mass of carbon dioxide formed `=1.76gm` Since 44 gm of carbon dioxide contains 12 gm carbon 1.76 gm of carbon dioxide contains 12 gm carbon 1.76 gm of carbon contains `=(12)/(44)xx1.76gm` carbon percentage of carbon `=(12)/(44)xx(1.76)/(0.92)xx100=52.17` percentage of oxygen`=100-`(percentage of `C+`percent of H) `=100-(52.17+13.04)` `100-65.21=34.79` Composition of the compound : `C=52.17%`, `H=13.04%`, `O=34.79%` |
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| 31. |
A compound contains `C=90%` and `H=10%` Empirical formula of the compound is:A. `C_(15)H_(30)`B. `C_(15)H_(20)`C. `C_3H_4`D. `C_3H_(10)` |
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Answer» Correct Answer - C `{:(C,,H),((90)/(12),,(10)/(1)),(7.5,,10),(1,,1.3),(3,,4):}` (whole number) `Ef=C_3H_4` |
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| 32. |
A compound contains `C=40%,O=53.5%`, and `H=6.5%` the empirical formula formula of the compound is:A. `CH_2O`B. `C_2H_4O`C. `C_6H_(12)O_6`D. `C_2H_4O_2` |
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Answer» Correct Answer - A `{:(C,,H,,O),((40)/(12),,(6.5)/(1),,(53.5)/(16)),(3.33,,6.5,,3.34),(1,,2,,1):}` `EF=CH_2O` |
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| 33. |
`n-`Butane `(C_4H_(10))` is produced by monobromination of `C_2H_6` followed by Wurtz reaction. Calculate the volume of ethane. The bromination takes place with `90%` yeild and the wurtz reaction with `85%` yield.A. 27.75 litresB. 55.5 litresC. 111 litresD. 5.55 litres |
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Answer» Correct Answer - B `2C_2H_6+Br_2rarr2C_2H_5Br+HBroverset(2Na)toC_4H_(10)` `12xx4+10=58gm` 58 gm of n-butene `implies2xx22.4` litre of `C_2H_6` at STP i.e., `2xx22.4xx(100)/(85)xx(100)/(90)` `=58.56` litre at STP `58gm` of n-butate`=58.56` litre `55gm` of n-butane`=55.5` litre |
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| 34. |
A mixture of formic acid and oxalic acid is heated with conc. `H_2SO_4` . The gas produced is collected and treated with `KOH` solution where the volume decreases by 1/6th. The molar ratio of two acids (formic acid/oxalic acid) in the original mixuture is: (a) `4:1` (b) `1:4` (c) `2:1` (d) `1:2` |
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Answer» Let `x` mol of `HCOOH` and y mole of `(COOH)_2` `underset(x)HCOOHoverset(H_2SO_4)toH_2O+underset(x)CO` `underset(y)(COOH)_2overset(H_2SO_4)toH_2O+underset(y)CO+underset(y)CO_2` total mol of `(CO+CO_2)=x+2y` Total mol of `CO_2=y` Accordig to the question, `(x+2y)xx(1)/(6)=y` `therefore(x)/(y)=4:1` Alternatively Mol fraction of `CO_2=(y)/(x+2y)=(1)/(6)` `(x)/(y)=4:1` |
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