1

### 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)
2

### 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.
3

### 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
4

### AIPMT 2011 Prelims

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