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Chemistry MCQs for NEET — Practice Questions with Answers

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Solution of sucrose (Mol. Mass = 342) is prepared by dissolving 34.2 gm. of it in 1000 gm. of water. freezing point of the solution is (K_f for water is 1.86 K kg mol^–1) [AIEEE 2003]

272.814 K

278.1 K

273.15 K

270 K

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Explanation

Molality of the solution $= \frac{34 .2}{342} = 0 .1$

$Δ T_{f} = K_{f} \times m = 1 .86 \times 0 .1 = 0 .186 K$

Freezing point of solution $= 273 - 0 .186 = 272 .814 K$

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An aqueous solution of a weak monobasic acid containing 0.1 g in 21.7 g of water freezes at 272.817K. If the value of K_f for water is 1.86 K kg mol^–1, the molecular mass of the acid is [AMU 2002]

46.0

48.6

48.8

46.8

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Explanation

Mass of solvent $(W_{A}) = 21 .7 g$

Mass of solute $(W_{B}) = 0 .1 g$

Depression in freezing point, $(Δ T_{f}) = 273 - 272 .817 = 0 .183 K$

$Δ T_{f} = k_{f} \times m = k_{f} . \frac{W_{B}}{W_{A}} \times \frac{1000}{M_{B}}$

$M_{B} = \frac{k_{f} \times W_{B} \times 1000}{W_{A} \times Δ T_{f}} = \frac{1 .86 \times 0 .1 \times 1000}{21 .7 \times 0 .183} = 46 .8$

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What is the molality of solution of a certain solute in a solvent if there is a freezing point depression of 0.184° and if the freezing point constant is 18.4

0.01

0.001

100

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Explanation

$Δ T_{f} = K_{f} \times m$

or $m = \frac{Δ T_{f}}{K_{f}} = \frac{0 .184}{18 .4} = 0 .01$

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A solution containing 6.8 g of a nonionic solute in 100 g of water was found to freeze at $- 0 {.93}^{o} C$ . The freezing point depression constant of water is 1.86. Calculate the molecular weight of the solute

[Pb. PMT 1994; ISM Dhanbad 1994]

13.6

136

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Explanation

$M_{B} = \frac{1000 \times K_{f} \times W_{B}}{Δ T_{f} \times W_{A}} = \frac{1000 \times 1 .86 \times 6 .8}{100 \times 0 .93} = 136$

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The molar freezing point constant for water is 1.86°C/mole. If 342 g of cane sugar $(C_{12} H_{22} O_{11})$ is dissolved in 1000 g of water, the solution will freeze at

–1.86°C

1.86°C

–3.92°C

2.42°C

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Explanation

$Δ T_{f} = 1 .86 (\frac{342}{342}) = 1 .86$

$Δ T_{f} = - 1 {.86}^{o} C$

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The molal freezing point constant for water is 1.86°C/m. Therefore, the freezing point of 0.1 m NaCl solution in water is expected to be [MLNR 1994]

–1.86°C

–0.186°C

–0.372°C

+0.372°C

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Explanation

$Δ T_{f} = i K_{f} m = 2 \times 1 .86 \times 0 .1 = 0 .372$

$T_{f} = - 0 {.372}^{o} C$

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The depression in freezing point of 0.01 M aqueous solutions of urea, sodium chloride and sodium sulphate is in the ratio of [Roorkee 1990; DCE 1994]

1 : 1 : 1

1 : 2 : 3

1 : 2 : 4

2 : 2 : 3

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Explanation

Concentration of particles of 0.01 M urea, NaCl and $N a_{2} S O_{4} = 0 .01 M, 0 .02 M, 0 .03$ respectively i.e., they are in the ratio 1 : 2 : 3. Hence, depression in freezing point will be in the same ratio.

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The Van't Hoff factor for 0.1 M $Ba {({NO}_{3})}_{2}$ solution is 2.74. The degree of dissociation is [IIT 1999]

91.3%

87%

100%

74%

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Explanation

$\underset{At. eq. (0 .1 - x)}{\underset{Initial 0 .1}{B a {(N O_{3})}_{2}}}$ â‡Œ $\underset{x M}{\underset{0}{B a^{+ +}}} + \underset{2 x m}{\underset{0}{2 N O_{3}^{-}}}$

$i = \frac{(0 .1 - x) + x + 2 x}{0 .1} = \frac{0 .1 + 2 x}{0 .1} = 2 .74 = 0 .1 + 2 x = 0 .274$ or $x = \frac{0 .174}{2} = 0 .087$

$% α = \frac{x}{0 .1} \times 100 = \frac{0 .087}{0 .1} \times 100 = 87 %$

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Assertion : Perfectly ideal solution is not possible with respect to binary solution of two liquids.

Reason : No two substances can have exactly the same nature of intermolecular forces & also of the same magnitude.

A. If both the assertion and the reason are true and the reason is a correct explanation of the assertion B. If both the assertion and reason are true but the reason is not a correct explanation of the assertion C. If the assertion is true but the reason is false D. If both the assertion and reason are false

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Explanation

(A) Assertion is true, Reason is true & Reason is the correct explanation of the Assertion.

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Assertion : When a cell is placed in hypertonic solution, it shrinks.

Reason : Reverse osmosis is used for desalination of water.

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Explanation

(B) Assertion is correct because fluids present inside the cell come out due to higher vapour pressure

inside the cell than outside the cell.

Reason is correct because in reverse osmosis, solvent from saline water enters the pure solvent

through semi-permeable membrane.

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