When 3 g of a non-volatile solute is dissolv | vedclass

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(D) Given: Mass of solute $(w_2)$ = $3 \ g$, Volume of water = $200 \ mL$, so Mass of solvent $(w_1)$ = $200 \ g = 0.2 \ kg$. Boiling point of solutio

Vedclass · Accepted Answer

$17.3$

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(D) Given: Mass of solute $(w_2)$ = $3 \ g$, Volume of water = $200 \ mL$, so Mass of solvent $(w_1)$ = $200 \ g = 0.2 \ kg$. Boiling point of solution $(T_b)$ = $100.52 \ ^oC$, Boiling point of pure water $(T_b^o)$ = $100 \ ^oC$. Elevation in boiling point $(\Delta T_b)$ = $T_b - T_b^o = 100.52 - 100 = 0.52 \ K$. Formula: $\Delta T_b = K_b 	imes m = K_b 	imes \frac{w_2 	imes 1000}{M_2 	imes w_1}$. Substituting values: $0.52 = 0.6 	imes \frac{3 	imes 1000}{M_2 	imes 200}$. $0.52 = 0.6 	imes \frac{15}{M_2}$. $M_2 = \frac{0.6 	imes 15}{0.52} = \frac{9}{0.52} \approx 17.3 \ g \ mol^{-1}$.

When $3 \ g$ of a non-volatile solute is dissolved in $200 \ mL$ of water, the boiling point of the solution becomes $100.52 \ ^oC$. If $K_b$ for water is $0.6 \ K \ kg \ mol^{-1}$, the molar mass of the solute is ......... $g \ mol^{-1}$.

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