1

### AIPMT 2012 Mains

Vapour pressure of chloroform (CHCl3) and dichloromethane (CH2Cl2) at 25oC are 200 mm Hg and 41.5 mm Hg respectively. Vapour pressure of the solution obtained by mixing 25.5 g of CHCl3 and 40 g of CH2Cl2 at the same temperature will be
(Molecular mass of CHCl3 = 119.5 u and molecular mass of CH2Cl2 = 85 u)
A
173.9 mm Hg
B
615.0 mm Hg
C
90.6 mm Hg
D
285.5 mm Hg

## Explanation

nCHCl3 = ${{25.5} \over {119.5}}$ = 0.213

nCH2Cl2 = ${{40} \over {85}}$ = 0.47

PT = PoAXA + PoBXB

= $200 \times {{0.213} \over {0.683}} + 41.5 \times {{0.47} \over {0.683}}$

= 62 + 28.55

= 90.63
2

### AIPMT 2012 Prelims

pA and pB are the vapour pressure of pure liquid components, A and B, respectively of an ideal binary solution. If xA represents the mole fraction of component A, the total pressure of the solution will be
A
${p_A} + {x_A}\left( {{p_B} - {p_A}} \right)$
B
${p_A} + {x_A}\left( {{p_A} - {p_B}} \right)$
C
${p_B} + {x_A}\left( {{p_B} - {p_A}} \right)$
D
${p_B} + {x_A}\left( {{p_A} - {p_B}} \right)$

## Explanation

p = pAxA + pBxB

= pAxA + pB(1 – xA)

(As for binary solution xA + xB = 1)

= pAxA + pB – pBxA

= pB + xA(pA – pB)
3

### AIPMT 2011 Prelims

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

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