1

### 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

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

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

${k_b} = 2{k_f}$
2

### 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
3

### JEE Main 2019 (Online) 11th January Morning Slot

The freezing point of a diluted milk sample is found to be –0.2oC, while it should have been –0.5oC for pure milk. How much water has been added to pure milk to make the diluted sample?
A
1 cup of water to 2 cups of pure milk
B
2 cups of water to 3 cups of pure milk
C
3 cups of water to 2 cups of pure milk
D
1 cup of water to 3 cups of pure milk

## Explanation

We know,

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

$\therefore$ $\Delta$TfDil. Milk = (1) $\times$ kf $\times$ mdil = 0.2 ...(1)

$\Delta$TfPure Milk = (1) $\times$ kf $\times$ mpure = 0.5 ...(2)

From (1) and (2), we get

${{{m_{pure}}} \over {{m_{dil}}}} = {{0.5} \over {0.2}}$ = ${5 \over 2}$ ....(3)

Assume moles of fat in both the milk samples = n

Weight of milk = w1

and weight of water = w0

Molality of pure milk,
mpure = ${n \over {{w_1}}} \times 1000$ ....(4)

Molality of diluted milk,
mdil. = ${n \over {{w_1} + {w_0}}} \times 1000$ ....(5)

From (4) and (5), we get

${{{m_{pure}}} \over {{m_{dil}}}} = {{{w_1} + {w_0}} \over {{w_1}}}$

Using equation (3), we get

${5 \over 2} = {{{w_1} + {w_0}} \over {{w_1}}}$

$\Rightarrow$ 5w1 = 2w1 + 2w0

$\Rightarrow$ 3w1 = 2w0

$\Rightarrow$ ${{{w_1}} \over {{w_0}}} = {2 \over 3}$

Which means 3 cups of water has been added to 2 cups of pure milk.
4

### JEE Main 2019 (Online) 11th January Evening Slot

K2Hgl4 is 40% ionised in aqueous solution. The value of its van't Hoff factor (i) is:
A
1.6
B
2.2
C
2.0
D
1.8

## Explanation

K2Hgl4 is 40% ionised.

$\therefore$ $\alpha$ = ${{40} \over {100}}$ = 0.4

K2[Hgl4] $\to$ 2K+ + [Hgl4]2+

N = ${{2 + 1} \over 1}$ = 3

i = 1 + (N - 1)$\alpha$

= 1 + (3 - 1)0.4

= 1 + 2$\times$0.4

= 1.8