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 $A$ is ________ $g / mol$ . (Round off to the Nearest Integer).

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Solution

Step 1: Calculate the mass of solvent for the solution of $A$
Mass of solution containing $A = 100 g$
Mass of solute $A = 18.1 g$
Thus, the mass of solvent for the solution of $A = 100 - 18.1 = 81.9 g$
Step 2: Calculate the molar mass of $A$
Let MM be the molar mass of A.
Since it is given that the boiling point of both the solutions is equal. Thus, elevation in boiling point is also equal for both solutions. So,
${(∆ T_{b})}_{K_{4} F e {(C N)}_{6}} = {(∆ T_{b})}_{A} {(i K_{b} m)}_{K_{4} F e {(C N)}_{6}} = {(i K_{b} m)}_{A} (1 + 4 α) \times 1 = 1 \times \frac{\frac{18.1}{M M}}{\frac{81.9}{1000}} (1 + (4 \times 0.4)) = \frac{18.1 \times 1000}{81.9 \times M M} \Rightarrow M M = 85 g / m o l$
Thus, the molar mass of $A$ is $85 g / m o l$ .

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A 1 molal K4Fe(CN)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 is ________ g/mol. (Round off to the Nearest Integer).

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