The correct relation between elevation of boiling point and molar mass of solute is:

M_{2} = \frac{K_{b} \times W_{2}}{Δ T_{b} \times W_{1}}

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M_{2} = \frac{K_{b} \times W_{1}}{Δ T_{b} \times W_{2}}

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M_{2} = \frac{Δ T_{b} \times K_{b}}{W_{1} \times W_{2}}

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M_{2} = \frac{Δ T_{b} \times W_{1}}{K_{b} \times W_{2}}

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Solution

The correct option is A $M_{2} = \frac{K_{b} \times W_{2}}{Δ T_{b} \times W_{1}}$
The correct relation between elevation of boiling point ( $Δ T_{b}$ ) and molar mass of solute ( $M_{2}$ ) is $M_{2} = \frac{K_{b} \times W_{2}}{Δ T_{b} \times W_{1}}$

Here, $W_{1}$ and $W_{2}$ represents masses of solvent and solute respectively. $K_{b}$ represents molal elevation in boiling point constant.

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Similar questions

Column I	Column II
(A) $K_{b}$	(i) $\frac{K_{b} \times W_{2} \times 1000}{{Δ T_{b} \times W}_{1}}$
(B) $M_{2}$	(ii) $\frac{W_{2} \times 1000}{{M_{2} \times W}_{1}}$
(C) $π$	(iii) $\frac{R T_{b}^{2}}{1000 \times L_{V}}$
(D) $m$	(iv) $\frac{Δ T_{b} \times d R T}{1000 \times K_{b}}$