QJEEM19Ch00a3: JEE Main 2019 Chemistry - Class 12 JEEM19Ch00a3 Solution: A solution containing 62 g ethylene glycol in 250 g water is…

Here water is solvent and ethylene glycol is solute.

We know, Depression of freezing point,

$\Delta$ T_f = K_f . m

$\Delta {T_f} = 1.86 \times {{\left( {{{62} \over {62}}} \right)} \over {\left( {{{250} \over {1000}}} \right)}}$

= 7.44

We know, freezing point of pure water (here solvent) is 0^oC. Which is represented by $T_f^0$ .

We also know, $\Delta {T_f} = T_f^0 - {T_f}$

where $T_f^0$ = Freezing point of pure solvent and ${T_f}$ = Freezing point of solution

$\therefore$ 7.44 = 0 - ${T_f}$

$\Rightarrow$ ${T_f}$ = -7.44^oC

So, not a single drop of water solution will become ice until temperature reaches -7.44^oC. When temperature decrease more than -7.44^oC then some part of the solution starts becoming ice staring from surface of the solution.

Now let W_l gm of solution still stays in liquid phase when temperature reaches -10^oC.

$\therefore$ $10 = 1.86 \times {{\left( {{{62} \over {62}}} \right)} \over {\left( {{{{W_l}} \over {1000}}} \right)}}$

W_l = 186 gm

So, The amount of water (in g) separated as ice is

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