1

### JEE Main 2019 (Online) 10th January Morning Slot

Liquids A and B form an ideal solution in the entire composition range. At 350 K, the vaapor pressures of pure A and pure B are 7 $\times$ 103 Pa and 12 $\times$ 103 Pa, respectively . The composition of the vapor in equilibriumwith a solution containing 40 mole percent of A at this temperature is :
A
xA = 0.76; xB = 0.24
B
xA = 0.28; xB = 0.72
C
xA = 0.4; xB = 0.6
D
xA = 0.37; xB = 0.63

## Explanation

${y_A} = {{{P_A}} \over {{P_{Total}}}} = {{P_A^0{x_A}} \over {P_A^0{X_A} \times P_B^0X{}_B}}$

$= {{7 \times {{10}^3} \times 0.4} \over {7 \times {{10}^3} \times 0.4 + 12 \times {{10}^3} \times 0.6}}$

$= {{2.8} \over {10}} = 0.28$

${y_B} = 0.72$
2

### JEE Main 2019 (Online) 10th January Evening Slot

The amount of sugar (C12H22O11) required to prepare 2L of its 0.1 M aqueous solution is :
A
17.1 g
B
34.2 g
C
68.4 g
D
136.8 g

## Explanation

Molarity = ${{{{(n)}_{solute}}} \over {{V_{solution}}(in\,\,lit)}}$

0.1 = ${{wt./342} \over 2}$

wt (C12H22O11) = 68.4 gram
3

### JEE Main 2019 (Online) 10th January Evening Slot

Elevation in the boiling point for 1 molar solution of glucose is 2 K. The depression in the freezing point for 2 molal solution of glucose in the same solvent is 2 K. The relation between Kb and Kf is
A
Kb = Kf
B
Kb = 0.5 Kf
C
Kb = 1.5 Kf
D
Kb = 2 Kf

## Explanation

${{\Delta {T_b}} \over {\Delta {T_f}}} = {{i.m \times {k_b}} \over {i \times m \times {k_f}}}$

${2 \over 2} = {{1 \times 1 \times {k_b}} \over {1 \times 2 \times {k_f}}}$

${k_b} = 2{k_f}$
4

### JEE Main 2019 (Online) 10th January Evening Slot

Haemoglobin and gold sol are examples of :
A
positively charged sols
B
negatively charged sols
C
Positively and negatively charged sols, respectively
D
negatively and positively charged sols, respectively

## Explanation

Haemoglobin $\buildrel \, \over \longrightarrow$ positive sol
Ag $-$ sol $\buildrel \, \over \longrightarrow$ negative sol