1

### JEE Main 2019 (Online) 9th January Evening Slot

A solution containing 62 g ethylene glycol in 250 g water is cooled to $-$ 10oC. If Kf for water is 1.86 K kg mol$-$1 , the amount of water (in g) separated as ice is :
A
48
B
32
C
64
D
16

## Explanation

$\Delta$Tf = Kf . m

$\Rightarrow$ Tf = Kf $\times$ ${w \over {M{W_{kg}}}}$

$\Rightarrow$10 = 1.86 $\times$ ${{62/62} \over {{W_{kg}}}}$

W = 0.186 kg

$\Delta$W = (250 $-$ 186) = 64 gm
2

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

A mixture of 100 m mol of Ca(OH)2 and 2 g of sodium sulphate was dissolved in water and the volume was made up to 100 mL. The mass of calcium sulphate formed and the concentration of OH in resulting solution, respectively, are : (Molar mass of Ca (OH)2, Na2SO4 and CaSO4 are 74, 143 and 136 g mol–1 , respectively; Ksp of Ca(OH)2 is 5.5 × 10–6 )
A
13.6g, 0.28 mol L$-$1
B
13.6g, 0.14 mol L$-$1
C
1.9g, 0.28 mol L$-$1
D
1.9g, 0.14 mol L$-$1

## Explanation

Ca(OH)2 + Na2SO4 $\to$ CaSO4 + 2NaOH

100 m mol 14 m mol      $-$     $-$      $-$

$-$    $-$      $-$  14 m mol 28 m mol

wCasO4 = 14 $\times$ 10$-$3 $\times$ 136 = 1.9 gm

[OH$-$] = ${{28} \over {100}}$ = 0.28 M
3

### 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

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

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

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

${y_B} = 0.72$
4

### 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