6 g of a mixture of naphthalene (C_{10}H_8) | vedclass

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(B) Given,weight of solvent $(W_A) = 300 \ g = 0.3 \ kg$. Depression in freezing point,$\Delta T_f = 0.70 \ K$. Molal depression constant,$K_f = 5.1 \

Vedclass · Accepted Answer

$3.40, 2.60$

Vedclass · Answer

(B) Given,weight of solvent $(W_A) = 300 \ g = 0.3 \ kg$. Depression in freezing point,$\Delta T_f = 0.70 \ K$. Molal depression constant,$K_f = 5.1 \ K \ kg \ mol^{-1}$. Using the formula $\Delta T_f = K_f 	imes m$,where $m$ is molality: $0.70 = 5.1 	imes \frac{	ext{Total moles of solute}}{0.3}$. Total moles of solute $= \frac{0.70 	imes 0.3}{5.1} = 0.04117 \ mol$. Let $x$ be the mass of naphthalene $(C_{10}H_8, 	ext{molar mass} = 128 \ g/mol)$ and $(6-x)$ be the mass of anthracene $(C_{14}H_{10}, 	ext{molar mass} = 178 \ g/mol)$. $\frac{x}{128} + \frac{6-x}{178} = 0.04117$. $0.0078125x + 0.033707 - 0.005618x = 0.04117$. $0.0021945x = 0.007463$. $x \approx 3.40 \ g$ (naphthalene). Mass of anthracene $= 6 - 3.40 = 2.60 \ g$. Thus,the composition is $3.40 \ g$ naphthalene and $2.60 \ g$ anthracene.

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