1

JEE Main 2017 (Online) 9th April Morning Slot

A solution is prepared by mixing 8.5 g of CH2Cl2 and 11.95 g of CHCl3 . If vapour pressure of CH2Cl2 and CHCl3 at 298 K are 415 and 200 mmHg respectively, the mole fraction of CHCl3 in vapour form is : (Molar mass of Cl = 35.5 g mol−1)
A
0.162
B
0.675
C
0.325
D
0.486
2

JEE Main 2018 (Offline)

For 1 molal aqueous solution of the following compounds, which one will show the highest freezing point?
A
[Co(H2O)3 Cl3].3H2O
B
[Co(H2O)6] Cl3
C
[Co(H2O)5 Cl] Cl2.H2O
D
[Co(H2O)4 Cl2] Cl.2H2O

Explanation

We know,

$\Delta {T_f}\, = \,im{K_f}$

Where

$\Delta {T_f}\, = \,$ Depression in freezing point

i = no of ions or molecule

m = molality of solute

Kf = freezing point depression constant.

In this question m = 1 and Kf is constant.

So, $\Delta {T_f}\,\, \propto \,\,i$

Which solutions have more i, those solutions $\Delta {T_f}\,$ will be more.

Now,

Freezing point of a aqueous solution ${f_p} = \, - \,\Delta {T_f}$

So, those solutions have more $\Delta {T_f}$ will have less freezing point

and $\Delta {T_f}$ will be more when value of i is more.

Now, for,

[ Co (H2O)6] Cl3 $\rightleftharpoons$ [ Co(H2O)6]3+ + 3Cl$-$

Value of i = 4 (no. of ions), 3 ions of Cl$-$

and one ion of [ Co (H2O)6]3+

[ Co (H2O)5 Cl ] Cl2 . H2O $\rightleftharpoons$ [ Co (H2O)5 Cl ]2+ + 2Cl$-$

Value of i = 3

[ Co (H2O)4 Cl2 ] Cl . 2H2O $\rightleftharpoons$ [ Co (H2O)4 Cl2 ] + + Cl$-$

Value of i = 2

and [ Co(H2O)3 Cl ] . 3H2O do not show ionization.

So, i = 1.

As, [ Co (H2O)3 Cl ]. 3H2O have lowest value of i,

So, it's freezing point will be maximum.
3

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

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

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