The addition of 0.643 ,g of a compound to 50 | vedclass

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(A) Given: Weight of solute $(w_2)$ = $0.643 \,g$ Freezing point constant $(K_f)$ = $5.12 \,K \,kg \,mol^{-1}$ Depression in freezing point $(\Delta T

Vedclass · Accepted Answer

$156$

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(A) Given: Weight of solute $(w_2)$ = $0.643 \,g$ Freezing point constant $(K_f)$ = $5.12 \,K \,kg \,mol^{-1}$ Depression in freezing point $(\Delta T_f)$ = $5.51^{\circ}C - 5.03^{\circ}C = 0.48 \,K$ Volume of benzene = $50 \,mL$ Density of benzene = $0.879 \,g \,mL^{-1}$ Weight of solvent $(w_1)$ = $50 \,mL 	imes 0.879 \,g \,mL^{-1} = 43.95 \,g$ Using the formula for molar mass $(M_2)$: $M_2 = \frac{K_f 	imes w_2 	imes 1000}{\Delta T_f 	imes w_1}$ $M_2 = \frac{5.12 	imes 0.643 	imes 1000}{0.48 	imes 43.95}$ $M_2 = \frac{3292.16}{21.096} \approx 156.05 \,g \,mol^{-1}$ Thus,the molar mass is approximately $156 \,g \,mol^{-1}$.

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