1 g of non-volatile non-electrolyte solute i | vedclass

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(C) The elevation in boiling point is given by the formula: $\Delta T_b = K_b 	imes m$,where $K_b$ is the ebullioscopic constant and $m$ is the molal

Vedclass · Accepted Answer

$1:5$

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(C) The elevation in boiling point is given by the formula: $\Delta T_b = K_b 	imes m$,where $K_b$ is the ebullioscopic constant and $m$ is the molality of the solution.Since the mass of the solute $(1 \ g)$ and the mass of the solvent $(100 \ g)$ are the same for both solvents $A$ and $B$,the molality $m$ is the same for both solutions $(m_A = m_B)$.Therefore,the ratio of the elevation in boiling points is directly proportional to the ratio of their ebullioscopic constants:$\frac{\Delta T_b (A)}{\Delta T_b (B)} = \frac{K_{b(A)}}{K_{b(B)}}$.Given that the ratio of the ebullioscopic constants is $\frac{K_{b(A)}}{K_{b(B)}} = \frac{1}{5}$,the ratio of the elevation in boiling points is $1:5$.

$1 \ g$ of non-volatile non-electrolyte solute is dissolved in $100 \ g$ of two different solvents $A$ and $B$ whose ebullioscopic constants are in the ratio of $1 : 5$. The ratio of the elevation in their boiling points,$\frac{\Delta T_b (A)}{\Delta T_b (B)}$ is

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