JEE Main 2021 — Solutions Question with Solution

Effective molality = 0.6 + 1.6 + 0.4 = 2.6 m

As elevation in boiling point is a colligative property which depends on the amount of solute. So, to have same boiling point, the molality of two solutions should be same.

Molality of non-electrolyte solution = molality of $$K_4[Fe(CN{)}_6]$$ = 2.6 m

Now, 18.1 weight per cent solution means 18.1 g solute is present in 100 g solution and hence, (100 $$-$$ 18.1) = 81.9 g water.

$$Molality=\frac{(Massofsolute/Molarmassofsolute)}{Massofsolventinkg}\times 1000$$

Now, $$2.6=\frac{18.1/M}{81.9/1000}$$

where, M is the molar mass of non-electrolyte solute

Molar mass of solute, M = 85