JEE Main 2019 — Solutions Question with Solution

Here water is solvent and ethylene glycol is solute.

We know, Depression of freezing point,

$$\Delta$$T_f = K_f . m

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

= 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.

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

W_l = 186 gm

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

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