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1

AIPMT 2011 Mains

MCQ (Single Correct Answer)
A 0.1 molal aqueous solution of a weak acid is 30% ionized. If Kf for water is 1.86oC/m, the freezing point of the solution will be
A
$$-$$0.18oC
B
$$-$$0.54oC
C
$$-$$0.36oC
D
$$-$$0.24oC

Explanation

We know that $$\Delta $$Tf = i × Kf × m

Here i is van’t Hoff’s factor.

i for weak acid is 1 + $$\alpha $$.

Here $$\alpha $$ is degree of dissociation i.e., 30/100 = 0.3

$$ \therefore $$ i = 1 + $$\alpha $$ = 1 + 0.3 = 1.3

$$ \therefore $$ Tf = i × Kf × m = 1.3 × 1.86 × 0.1 = 0.24

$$ \therefore $$ Freezing point = – 0.24
2

AIPMT 2011 Prelims

MCQ (Single Correct Answer)
The freezing point depression constant for water is $$-$$1.86o C m$$-$$1. If 5.00 g Na2SO4 is dissolved in 45.0 g H2O, the freezing point is changed by $$-$$3.82oC. Calculate the van't Hoff factor for Na2SO4.
A
2.05
B
2.63
C
3.11
D
0.381

Explanation

According to depression in freezing point,

$$\Delta $$Tf = i $$ \times $$ Kf $$ \times $$ m

= i $$ \times $$ Kf $$ \times $$ $${{{w_B} \times 1000} \over {{m_B} \times {w_A}}}$$

Given : $$\Delta $$Tf = 3.82, Kf = 1.86,

wB = 5, mB = 142, wA = 45

i = $${{\Delta T \times {m_B} \times {w_A}} \over {{K_f} \times {w_B} \times 1000}}$$

= $${{3.82 \times 142 \times 45} \over {1.86 \times 5 \times 1000}}$$

= 2.63
3

AIPMT 2011 Prelims

MCQ (Single Correct Answer)
The van't Hoff factor $$i$$ for a compound which undergoes dissociation in one solvent and association in other solvent is respectively
A
less than one and greater than one
B
less than one and less than one
C
greater than one and less than one
D
greater than one and greater than one.

Explanation

From the value of van’t Hoff factor i it is possible to determine the degree of dissociation or association. In case of dissociation, i is greater than 1 and in case of association i is less than 1.
4

AIPMT 2011 Mains

MCQ (Single Correct Answer)
200 mL of an aqueous solution of a protein contains its 1.26 g. The osmotic pressure of this solution at 300 K is found to be 2.57 $$ \times $$ 10$$-$$3 bar. The molar mass of protein will be (R = 0.083 L bar mol$$-$$1 K$$-$$1)
A
51022 g mol$$-$$1
B
122044 g mol$$-$$1
C
31011 g mol$$-$$1
D
61038 g mol$$-$$1

Explanation

Osmotic pressure, $$\pi $$ = CRT

$$ \Rightarrow $$ $$\pi $$ = $${n \over V}$$RT

$$ \Rightarrow $$ $$\pi $$V = $${w \over M}$$RT

M = $${{wRT} \over {\pi V}}$$

= $${{1.26 \times 0.083 \times 300} \over {2.57 \times {{10}^{ - 3}} \times {{200} \over {1000}}}}$$

= $${{1.26 \times 0.083 \times 300} \over {2.57 \times {{10}^{ - 3}} \times 0.2}}$$

= 61038 g mol–1

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