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 Multiple Choice QuestionsMultiple Choice Questions

1.

A 5.2 molal aqueous solution of methyl alcohol, CH3OH, is supplied. What is the mole fraction of methyl alcohol in the solution?

  • 0.100

  • 0.190

  • 0.086 

  • 0.050


C.

0.086 

Molality space equals space fraction numerator Moles space of space solute over denominator Mass space of space solvent space left parenthesis in space kg right parenthesis end fraction
space equals space fraction numerator 5.2 space mol space CH subscript 3 OH over denominator 1 space kg space left parenthesis equals space 1000 space straight g right parenthesis space straight H subscript 2 straight O end fraction
straight n subscript 1 left parenthesis CH subscript 3 OH right parenthesis space equals space 5.2
straight n subscript 2 space left parenthesis straight H subscript 2 straight O right parenthesis space equals space 1000 over 18 space equals space 55.56
straight n subscript 1 plus straight n subscript 2 space equals space 5.20 space plus space 55.66 space equals space 60.76 space mol
straight X subscript CH subscript 3 OH end subscript space space equals space Mole space fraction space of space CH subscript 3 OH
space equals space fraction numerator straight n subscript 1 over denominator straight n subscript 1 plus straight n subscript 2 end fraction space equals space fraction numerator 52 over denominator 60.76 end fraction space equals space 0.086
596 Views

2.

The density of a solution prepared by dissolving 120 g of urea (mol. Mass = 60 u ) in 1000g of water is 1.15 g/mL. The molarity of this solution is

  • 0.50 M

  • 1.78 M

  • 1.02 M

  • 2.05


D.

2.05

molarity space equals space fraction numerator moles space of space space solute over denominator volume space of space solution end fraction
Total mass of solution
= 1000 g water +120 g urea
 = 1120 g
Density space of space solution space space equals space 1.15 space straight g divided by ml
Thus comma space volume space of space solution space equals space mass over density space equals space fraction numerator 1120 space straight g over denominator 1.15 space straight g divided by mL end fraction
space equals space 973.91 space mL space equals space 0.974 space straight L
Molarity space equals space fraction numerator 2 over denominator 0.974 end fraction space equals space 2.05 space mol space straight L to the power of negative 1 end exponent
space equals space 20.5 space straight M
283 Views

3.

The molarity of a solution obtained by mixing 750 mL of 0.5 (M) HCl with 250 mL of 2(M) HCl will be

  • 0.875 M

  • 1.00 M

  • 1.75 M

  • 0.0975 M


A.

0.875 M

M1V1 + M2V2 = MV


straight M space equals fraction numerator straight M subscript 1 straight V subscript 1 space plus straight M subscript 2 straight V subscript 2 over denominator straight V end fraction space equals space fraction numerator 0.5 space straight x 750 plus 2 space straight x 250 over denominator 1000 end fraction
M= 0.875
215 Views

4.

3 g of activated charcoal was added to 50 mL of acetic acid solution (0.06N) in a flask. After an hour it was filtered and the strength of the filtrate was found to be 0.042 N. The amount of acetic acid adsorbed (per gram of charcoal) is:

  • 18 mg

  • 36 mg

  • 42 mg

  • 54 mg


A.

18 mg

The initial strength of acetic acid = 0.06N
Final strength = 0.042 N
Volume given = 50 mL
there Initial m moles of CH3COOH 
 = 0.06 x 50 = 3
Final m moles of CH3COOH 
 = 0.042 x 50 = 21
therefore, m moles of CH3COOH absorbed
 = 3-2.1
 = 0.9 m mol
Hence, mass of CH3COOH  absorbed per gram of charcoal
 = fraction numerator 0.9 space straight x space 60 over denominator 3 end fraction
54 over 3 space equals space 18 space mg

424 Views

5.

Consider separate solution of 0.500 M C2H5OH (aq), 0.100 M Mg3(PO4)2 (aq) 0.250 M KBr (aq) and 0.125 M Na3PO4 (aq) at 25oC. Which statement is true about these solutions, assuming all salts to be strong electrolytes?

  • They all have the same osmotic pressure

  • 0.100 M Mg3(PO4)2 (aq) has the highest osmotic pressure

  • 0.125 M Na3PO (aq) has the highest osmotic pressure

  • 0.500 M C2H5OH (aq) has the higest osmotic pressure


A.

They all have the same osmotic pressure

effective molarity = Van't Hoff factor x molarity
0.5 M C2H5OH (aq)      i =1
Effective molarity = 0.5

0.25 M KBr (aq)          i = 2
Effective molarity  = 0.5

0.1 M Mg3(PO4)2 (aq) i = 5
Effective molarity = 0.5 M

0.125 M Na3PO4 (aq)
Effective molarity = 0.5 M

Hence, all colligative properties are same

960 Views

6.

For the estimation of nitrogen 1.4g of an organic compound was digested by Kjeldahl's method and the evolved ammonia was absorbed in 60 mL of M/10 sulphuric acid. The unreacted acid required 20 mL of M/10 sodium hydroxide for complete neutralisation. The percentage of nitrogen in the compound is 

  • 6%

  • 10%

  • 3%

  • 5%


B.

10%

386 Views

7.

Kf for water is 1.86K kg mol–1.If your automobile radiator holds 1.0 kg of water, how many grams of ethylene glycol (C2H6O2) must you add to get the freezing point of the solution lowered to –2.8°C?

  • 72 g

  • 93 g

  • 39 g

  • 27 g


B.

93 g

Coolant is glycol (C2H6O2) and is non-electrolyte.
ΔTf =2.8°
increment straight T subscript straight f space equals space fraction numerator 1000 space straight x space straight K subscript straight f space straight x space straight w subscript 1 over denominator straight m subscript 1 straight w subscript 2 end fraction space rightwards double arrow
2.8 space equals space fraction numerator 1000 space straight x space 1.86 space straight x space straight w subscript 1 over denominator 62 space straight x space 1000 end fraction
therefore straight w subscript 1 space equals space 93.33 space straight g

314 Views

8.

The vapour pressure of acetone at 20oC is 185 torr. When 1.2 g of a non-volatile substance was dissolved in 100 g of acetone at 20oC, its vapour pressure was 183 torr. The molar mass (g mol-1 ) of the substance is:

  • 32

  • 64

  • 128

  • 488


B.

64

Given, 
po = 185 torr at 20oC
ps = 183 torr at 20oC
Mass of non-volatile substance,
m= 1.2 g
Mass of acetone taken = 100 g
As we have,fraction numerator straight p subscript straight o minus straight p subscript straight s over denominator straight p subscript straight s end fraction space equals space straight n over straight N
putting the values, we get


fraction numerator 185 minus 183 over denominator 183 end fraction space equals space fraction numerator begin display style fraction numerator 1.2 over denominator straight M end fraction end style over denominator begin display style 100 over 58 end style end fraction space
rightwards double arrow 2 over 183 space equals space fraction numerator 1.2 space straight x space 58 over denominator 100 space straight x space straight M end fraction
therefore space straight M space equals space fraction numerator 183 space straight x space 12 space straight x space 58 over denominator 2 space straight x space 100 end fraction
straight M space equals space 63.684 space almost equal to space 64 space straight g divided by mol

514 Views

9.

The ratio of masses of oxygen and nitrogen of a particular gaseous mixture is 1:4. The ratio of number of their molecule is

  • 1:4

  • 7:32

  • 1:8

  • 3:16


B.

7:32

straight n subscript straight O subscript 2 over straight n subscript straight N subscript 2 end subscript space equals space fraction numerator begin display style straight m subscript straight O subscript 2 end subscript over straight M subscript straight O subscript 2 end subscript end style over denominator begin display style straight m subscript straight N subscript 2 end subscript over straight M subscript straight N subscript 2 end subscript end style end fraction
where space straight m subscript straight O subscript 2 end subscript space equals space given space mass space of space straight O subscript 2
straight m subscript straight N subscript 2 end subscript space equals space given space mass space of space straight N subscript 2
straight M subscript straight O subscript 2 space end subscript space equals space molecular space mass space of space straight O subscript 2
straight M subscript straight N subscript 2 end subscript space equals space moleular space mass space space of space straight N subscript 2
straight n subscript straight O subscript 2 end subscript space equals space number space of space moles space of space straight O subscript 2
straight n subscript straight N subscript 2 space equals space end subscript Number space of space moles space of space straight N subscript 2
space equals space open square brackets straight m subscript straight O subscript 2 end subscript over straight m subscript straight N subscript 2 end subscript close square brackets 28 over 32 space equals space 1 fourth space straight x space 28 over 32 space equals space 7 over 32
231 Views

10.

18 g glucose (C6H12O6 ) is added to 178.2 g water. The vapor pressure of water (in torr) for this aqueous solution is:

  • 76.0

  • 752.4

  • 759.0

  • 7.6


B.

752.4

Vapour pressure of water (p°) = 760 torr

Number of moles of glucose
equals space fraction numerator Mas space left parenthesis straight g right parenthesis over denominator Molecular space mass space left parenthesis straight g space mol to the power of negative 1 end exponent right parenthesis end fraction

equals space fraction numerator 18 space straight g over denominator 180 space straight g space mol to the power of negative 1 end exponent end fraction space equals space 0.1
Molar mass of water = 18 g/mol
Mass of water (given) = 178.2 g

Number of moles of water


equals space fraction numerator Mass space of space water over denominator Molar space mass space of space water end fraction space equals space fraction numerator 178.2 over denominator 18 space straight g end fraction space equals space 9.9 space mol

Total number of moles = (0.1 +9.9) = 10 moles

Now, the mole fraction of glucose in solution = Change in pressure with respect to initial pressure.

straight i. straight e. space fraction numerator increment straight p over denominator increment straight p to the power of straight o end fraction space equals space fraction numerator 0.1 over denominator 10 end fraction
or space increment straight p space equals space 0.01 space straight p to the power of straight o space equals space 0.01 space straight x space 760 space equals space 7.6 space torr
therefore, Vapour pressure of solution = (760-7.6)torr
=752.4 torr


422 Views