When 20 g of naphthoic acid ($C_{11}H_{8}O_{2}$) is dissolved in 50 g of benzene ($k_{f} = 1.72\ K\ kg\ mol^{- 1}$), a freezing point depression of 2 K is observed. The van’t Hoff factor (i) is

Question

Dr. Abubacker · Accepted Answer

(a)
Actual molecular weight of naphthoic acid ($C_{11}H_{8}O_{2}$)=172
Molecular mass (calculated)
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  = \frac{1000 	imes k_{f} 	imes w}{W 	imes \mathrm{\Delta}T_{f}\ \ \ }\ \ \ \$
=$\frac{1000\  	imes 1.72 	imes \ 20}{50 	imes 2}$=344
van’t Hoff factor (i) =$\frac{actual\ mol.wt.}{calculated\ mo.wt.}\  = \frac{172}{344}\ \$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \  = 0.5$

Q. When 20 g of naphthoic acid ($C_{11}H_{8}O_{2}$) is dissolved in 50 g of benzene ($k_{f} = 1.72\ K\ kg\ mol^{- 1}$), a freezing point depression of 2 K is observed. The van’t Hoff factor (i) is

Explanation

Reviewed by Dr. Abubacker

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