QJEEM21Chdb1f: JEE Main 2021 Chemistry - Class 12 JEEM21Chdb1f Solution: A 1 molal K 4 Fe(CN) 6 solution has a degree of dissociation of …

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.

$Molality = {{(Mass\,of\,solute/Molar\,mass\,of\,solute)} \over {Mass\,of\,solvent\,in\,kg}} \times 1000$

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

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

Molar mass of solute, M = 85