A 1 molal K 4 Fe ( CN ) 6 solution has a degree of dissociation of 0 . 4   . Its boiling point is equal…

${\mathrm{K}}_4\mathrm{Fe}(\mathrm{CN}{)}_6 \rightleftharpoons 4{ \mathrm{K}}^{+} + \mathrm{Fe}(\mathrm{CN}{)}_6^{4-}$

$\begin{array}{lll} \mathrm{Initial} \mathrm{conc}. 1 \mathrm{m} & 0 & 0\end{array}$

$\begin{array}{llll}\text{ Final conc. }& (1-0.4)\mathrm{m} & 4\times 0.4\mathrm{m} & 0.4 \mathrm{m}\end{array}$

$=0.6 \mathrm{m} =1.6 \mathrm{m} =0.4 \mathrm{m}$

Effective molality $=0.6+1.6+0.4=2.6 \mathrm{m}$

For same boiling point, the molality of another solution should also be $2.6 \mathrm{m}$.

Now, $18.1$ weight percent solution means

$18.1\mathrm{gm}$ solute is present in $100\mathrm{gm}$ solution and hence, $(100-18.1=)81.9\mathrm{gm}$ water. Now, $2.6=\frac{18.1/\mathrm{M}}{81.9/1000}$

$\therefore$ Molar mass of solute, $\mathrm{M}=85$