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Avogadro's Law & Gay Lussac's Law

A 1 molal K4F...

Question

A 1 molal $K_{4} F e (C N)_{6}$ solution has a degree of dissociation of 0.4. Its boiling point is equal to that of another solution which contains 18.1 weight per cent of a nonelectrolyte solute A. The molar mass of A in $g / m o l$ is (Round off to the Nearest Integer)

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Solution

Answer: 85

Since boiling point is the same, elevation in boiling point is also the same for both solutions.
$i_{K_{4} [F e (C N)_{6}} = 1 + (n - 1) α$
Since it dissocates into 5 ions, n=5
$i_{K_{4} [F e (C N)_{6}} = 1 + 4 α$

$(Δ T_{B})_{K_{4} [F e (C N)_{6}]} = (Δ T_{B})_{A}$

${(i k_{B} m)}_{K_{4} [F e (C N)_{6}]} = (i k_{B} m)_{A}$

$k_{B}$ is same for both.
Also,
18.1 weight percent means 18.1 g of solute A is dissolved in 100 g of solution.
$Mass of solvent = (100 - 18.1) g$

$(1 + 4 α) \times 1 = \frac{1 \times \frac{18.1}{M} \times 1000}{100 - 18.1}$
Here M is the molar mass of solute A

$1 + (4 \times 0.4) = \frac{18.1}{M} \times \frac{1000}{81.9}$

$M = 85 g / m o l$

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