C_{6}H_{6} freezes at 5.5^{circ} C. The tempe | vedclass

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(B) Given: Mass of solute $(C_{4}H_{10})$ = $10 \ g$ Molar mass of $C_{4}H_{10}$ = $(4 	imes 12) + (10 	imes 1) = 58 \ g/mol$ Moles of solute = $\fr

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(B) Given: Mass of solute $(C_{4}H_{10})$ = $10 \ g$ Molar mass of $C_{4}H_{10}$ = $(4 	imes 12) + (10 	imes 1) = 58 \ g/mol$ Moles of solute = $\frac{10}{58} \ mol$ Mass of solvent $(C_{6}H_{6})$ = $200 \ g = 0.2 \ kg$ Molality $(m)$ = $\frac{	ext{moles of solute}}{	ext{mass of solvent in kg}} = \frac{10/58}{0.2} = \frac{10}{58 	imes 0.2} = \frac{10}{11.6} \approx 0.862 \ m$ Freezing point depression constant $(K_{f})$ = $5.12 \ ^{\circ} C/m$ Depression in freezing point $(\Delta T_{f})$ = $K_{f} 	imes m = 5.12 	imes 0.862 \approx 4.41 \ ^{\circ} C$ Freezing point of solution $(T_{f})$ = $T_{f}^{\circ} - \Delta T_{f} = 5.5 - 4.41 = 1.09 \ ^{\circ} C$ Rounding to the nearest integer,the temperature is $1 \ ^{\circ} C$.

$C_{6}H_{6}$ freezes at $5.5^{\circ} C$. The temperature at which a solution of $10 \ g$ of $C_{4}H_{10}$ in $200 \ g$ of $C_{6}H_{6}$ freezes is $...... \ ^{\circ} C$. (The molal freezing point depression constant of $C_{6}H_{6}$ is $5.12 \ ^{\circ} C/m$.)

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