Calculate the molar mass of a non-volatile so | vedclass

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(C) Step $1$: Calculate the elevation in boiling point $\Delta T_b = T_b - T_b^\circ = 319.4 	ext{ K} - 319 	ext{ K} = 0.4 	ext{ K}$.Step $2$: Use

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$240 	ext{ g mol}^{-1}$

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(C) Step $1$: Calculate the elevation in boiling point $\Delta T_b = T_b - T_b^\circ = 319.4 	ext{ K} - 319 	ext{ K} = 0.4 	ext{ K}$.Step $2$: Use the formula $\Delta T_b = K_b 	imes m$, where $m$ is molality.Step $3$: Molality $m = \frac{w_2 	imes 1000}{M_2 	imes w_1}$, where $w_2 = 4 	ext{ g}$, $w_1 = 100 	ext{ g}$, and $M_2$ is the molar mass of the solute.Step $4$: Substitute the values: $0.4 = 2.4 	imes \frac{4 	imes 1000}{M_2 	imes 100}$.Step $5$: Simplify: $0.4 = 2.4 	imes \frac{40}{M_2} \implies 0.4 = \frac{96}{M_2}$.Step $6$: Solve for $M_2$: $M_2 = \frac{96}{0.4} = 240 	ext{ g mol}^{-1}$.

Calculate the molar mass of a non-volatile solute when $4 \text{ g}$ of it is dissolved in $100 \text{ g}$ of a solvent that boils at $319.4 \text{ K}$. Given: $K_b = 2.4 \text{ K kg mol}^{-1}$ and the boiling point of the pure solvent is $319 \text{ K}$.

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