1

### 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.
2

### 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
3

### JEE Main 2019 (Online) 12th January Morning Slot

Freezing point of a 4% aqueous solution of X is equal to freezing point of 12% aqueous solution of Y. If molecular weight of X is A, then molecular weight of Y is -
A
4A
B
2A
C
3A
D
A

## Explanation

For same freezing point,

($\Delta$Tf)X = ($\Delta$Tf)Y

$\Rightarrow$ kf mx = kf my

$\Rightarrow$ ${{4 \times 1000} \over {A \times 96}} = {{12 \times 1000} \over {M \times 88}}$

$\Rightarrow$ M = 3.27A $\simeq$ 3A
4

### JEE Main 2019 (Online) 12th January Evening Slot

Molecules of benzoic acid (C6H5COOH) dimerise in benzene. 'w' g of the acid dissolved in 30 g of benzene shows a depression in freezing point equal to 2K. If the percentage association of the acid to form dimmer in the solution is 80, then w is – (Its given that Kf = 5 K kg mol–1, Molar mass of benzoic acid = 122 g mol–1)
A
1.5 g
B
1.8 g
C
1.0 g
D
2.4 g

## Explanation

We know,

$\Delta$Tf = i Kf $\times$ ${w \over M} \times {{1000} \over {{w_s}}}$

i = 1 + $\alpha$$\left( {{1 \over n} - 1} \right)$

Here Benzoic acid dimerise, so value of n = 2

$\therefore$  i = 1 + 0.8 $\left( {{1 \over 2} - 1} \right)$

= 1 $-$ 0.4

= 0.6

$\therefore$  2 = 0.6 $\times$ 5 $\times$ ${w \over {122}} \times {{1000} \over {30}}$

$\Rightarrow$  w = 2.44 g