A 1 molal K _ 4 Fe ( CN ) _ 6 solution has a degree of dissociation...

${K}_{4}Fe ( CN{)}_{6} ⇌ 4{K}^{+} + Fe ( CN{)}_{6}^{4 -}$

$\begin{cases}Initialconc . 1m & 0 & 0\end{cases}$

$\begin{cases}\text{Final conc.} & ( 1 - 0 . 4 ) m & 4 \times 0 . 4m & 0 . 4m\end{cases}$

$= 0 . 6m = 1 . 6m = 0 . 4m$

Effective molality $= 0 . 6 + 1 . 6 + 0 . 4 = 2 . 6m$

For same boiling point, the molality of another solution should also be $2 . 6m$.

Now, $18 . 1$ weight percent solution means

$18 . 1gm$ solute is present in $100gm$ solution and hence, $( 100 - 18 . 1 = ) 81 . 9gm$ water. Now, $2 . 6 = \dfrac{18 . 1 / M}{81 . 9 / 1000}$

$∴$ Molar mass of solute, $M = 85$