JEE Main · 2022 · 27 Jun 2022 · Shift 1 · Solutions

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2 g of a non-volatile non-electrolyte solute is dissolved in 200 g of two different solvents A and B whose ebullioscopic constants are in the ratio of

1 : 8

. The elevation in boiling points of A and B are in the ratio

\frac{x}{y} (x : y)

. The value of

y

is ______ (Nearest integer)

Answer: 8

Solution

The elevation in boiling point is given by

Δ T_{b} = K_{b} \cdot m

, where m is molality. Since the same amount of solute (2 g) is dissolved in the same amount of solvent (200 g) for both A and B, the molality is the same. Therefore,

\frac{( Δ T _{b} ) _{A}}{( Δ T _{b} ) _{B}} = \frac{( K _{b} ) _{A} \cdot m}{( K _{b} ) _{B} \cdot m} = \frac{( K _{b} ) _{A}}{( K _{b} ) _{B}} = \frac{1}{8} = \frac{x}{y}

. Thus, x = 1 and y = 8.

JEE Main · 2022 · 27 Jun 2022 · Shift 1 · Integer answer

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