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(D) Given:Mass of solute $(w_2)$ $= 2.5 	imes 10^{-3} \ kg = 2.5 \ g$Mass of solvent $(w_1)$ $= 75 	imes 10^{-3} \ kg = 75 \ g$$K_b = 0.52 \ K \ kg

Vedclass · Accepted Answer

$45$

Vedclass · Answer

(D) Given:Mass of solute $(w_2)$ $= 2.5 	imes 10^{-3} \ kg = 2.5 \ g$Mass of solvent $(w_1)$ $= 75 	imes 10^{-3} \ kg = 75 \ g$$K_b = 0.52 \ K \ kg \ mol^{-1}$Boiling point of solution $(T_b)$ $= 373.535 \ K$Boiling point of pure water $(T_b^o)$ $= 373.15 \ K$Elevation in boiling point $(\Delta T_b)$ $= T_b - T_b^o = 373.535 - 373.15 = 0.385 \ K$Formula: $\Delta T_b = K_b 	imes \frac{w_2 	imes 1000}{M_2 	imes w_1}$$0.385 = 0.52 	imes \frac{2.5 	imes 1000}{M_2 	imes 75}$$M_2 = \frac{0.52 	imes 2500}{0.385 	imes 75} = \frac{1300}{28.875} \approx 45.02 \ g \ mol^{-1}$Rounding to the nearest integer,the molar mass is $45 \ g \ mol^{-1}$.

$A$ solution containing $2.5 \times 10^{-3} \ kg$ of a solute dissolved in $75 \times 10^{-3} \ kg$ of water boils at $373.535 \ K$. The molar mass of the solute is $..... \ g \ mol^{-1}$. [nearest integer] (Given: $K_b(H_2O) = 0.52 \ K \ kg \ mol^{-1}$,boiling point of water $= 373.15 \ K$)

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