1

### JEE Main 2018 (Online) 15th April Evening Slot

Two 5 molal solutions are prepared by dissolving a non-electrolyte non-volatile solute separately in the solvents X and Y. The molecular weights of the solvents are Mx and My, respectively where Mx = ${3 \over 4}$ My. The relative lowering of vapor pressure of the solution in X is ''m'' times that of the solution in Y. Given that the number of moles of solute is very small in comparison to that of the solvent, the value of ''m'' is :
A
${4 \over 3}$
B
${3 \over 4}$
C
${1 \over 2}$
D
${1 \over 4}$

## Explanation

Relative lowering of vapour pressure,

${{\Delta P} \over P}$   =   ${{{n_2}} \over {{n_1}}}$

n2  =  Number of moles of solute

n1   =   Number of moles of solvent.

Given that,

Here is 5 molal solution, means 5 moles of solute are dissolved in 1 kg or 1000 g of solvent.

$\therefore\,\,\,$ Number of moles of solute = 5

Number of moles of solvent X  =  ${{1000} \over {{M_X}}}$

Number of moles of solvent Y = ${{1000} \over {My}}$

$\therefore\,\,\,$ ${\left( {{{\Delta \,P} \over P}} \right)_x}$   =   ${5 \over {{{1000} \over {{M_x}}}}}$ = ${{5{M_x}} \over {1000}}$

${\left( {{{\Delta \,P} \over P}} \right)_y}$   =   ${5 \over {{{1000} \over {{M_y}}}}}$   =   ${{5\,{M_y}} \over {1000}}$

Accoding to the question.

${{5{M_x}} \over {1000}}$   =   m $\times$ ${{5\,{M_y}} \over {1000}}$

$\Rightarrow $$\,\,\, Mx = m \times My \Rightarrow$$\,\,\,$ ${3 \over 4}$ My = m $\times$ My   [as given,   Mx = ${3 \over 4}$ My]

$\Rightarrow $$\,\,\, m = {3 \over 4} 2 MCQ (Single Correct Answer) ### JEE Main 2018 (Online) 16th April Morning Slot The mass of a non-volatile, non-electrolyte solute (molar mass = 50 g mol-1 ) needed to be dissolved in 114 g octane to reduce its vapour pressure to 75%, is : A 37.5 g B 75 g C 150 g D 50 g ## Explanation Molar mass of octane (C8 H18) = 8 \times 12 + 18 = 114 g/mol Let, \omega is the mass of solute. Relative lowering vapour pressure, {{\Delta P} \over P} = {{{\omega \over {50}}} \over {{\omega \over {50}} + {{114} \over {114}}}} \Rightarrow$$\,\,\,\,$ ${{75} \over {100}}$ = ${{{\omega \over {50}}} \over {{\omega \over {50}} + 1}}$

$\Rightarrow$ $\,\,\,\,$ ${3 \over 4}$ $\left( {{\omega \over {50}} + 1} \right) = {\omega \over {50}}$

$\Rightarrow$ $\,\,\,\,$ $\omega$ = 150 g
3

### 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
4

### 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