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(A) The boiling point elevation is given by the formula: $\Delta T_b = K_b 	imes m$,where $m$ is the molality of the solution.Molality $(m)$ is defin

Vedclass · Accepted Answer

$M_2 = \frac{1000 	imes K_b 	imes W_2}{\Delta T_b 	imes W_1}$

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(A) The boiling point elevation is given by the formula: $\Delta T_b = K_b 	imes m$,where $m$ is the molality of the solution.Molality $(m)$ is defined as the number of moles of solute $(n_2)$ per kilogram of solvent ($W_1$ in grams): $m = \frac{n_2 	imes 1000}{W_1}$.Since $n_2 = \frac{W_2}{M_2}$,where $W_2$ is the mass of solute and $M_2$ is the molar mass of solute,we substitute this into the molality equation:$m = \frac{W_2 	imes 1000}{M_2 	imes W_1}$.Substituting this into the boiling point elevation formula:$\Delta T_b = K_b 	imes \frac{W_2 	imes 1000}{M_2 	imes W_1}$.Rearranging for $M_2$,we get: $M_2 = \frac{1000 	imes K_b 	imes W_2}{\Delta T_b 	imes W_1}$.

What is the relation between the molar mass of a solute and the boiling point elevation?

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