2 text{ g} of a non-electrolyte solute, when | vedclass

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(C) Given:Mass of solute $(w_2)$ = $2 	ext{ g}$Mass of solvent $(w_1)$ = $50 	ext{ g} = 0.05 	ext{ kg}$Elevation in boiling point $(\Delta T_b)$ =

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$160 	ext{ g mol}^{-1}$

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(C) Given:Mass of solute $(w_2)$ = $2 	ext{ g}$Mass of solvent $(w_1)$ = $50 	ext{ g} = 0.05 	ext{ kg}$Elevation in boiling point $(\Delta T_b)$ = $0.13 	ext{ K}$Molal elevation constant $(K_b)$ = $0.52 	ext{ K kg mol}^{-1}$Formula: $\Delta T_b = K_b 	imes m = K_b 	imes \frac{w_2 	imes 1000}{M_2 	imes w_1 	ext{ (in g)}}$Rearranging for molar mass $(M_2)$: $M_2 = \frac{K_b 	imes w_2 	imes 1000}{\Delta T_b 	imes w_1}$$M_2 = \frac{0.52 	imes 2 	imes 1000}{0.13 	imes 50}$$M_2 = \frac{1040}{6.5} = 160 	ext{ g mol}^{-1}$

$2 \text{ g}$ of a non-electrolyte solute, when dissolved in $50 \text{ g}$ of water, increases its boiling point by $0.13 \text{ K}$. Calculate the molar mass of the solute if the molal elevation constant $(K_b)$ of water is $0.52 \text{ K kg mol}^{-1}$.

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