Question

2.0 g of benzoic acid dissolved in 25.0g of benzene shows a depression in freezing point equal to 1.62 K. Molal depression constant, $K_f$ of benzene is 4.9 K kg $mo{l}^{-1}$ , what is the percentage association of the acid if it forms dimer in the solution?

• A. 78.2%
• B. 82.6%
• C. 89.7%
• D. 99.2%

Answer 1

Answer: (D)

Mass of solute (benzoic acid), $w_2=2.0g$ Mass of solvent (benzene) $w_1=25.0g$ Observed $\Delta T_f=1.62K$ $k_f=4.9Kkgmo{l}^{-1}$ Observed molar mass of benzoic acid, $M_2=\frac{1000\times K_f\times w_2}{\Delta T_f\times w_1}$ $=\frac{1000\times 4.9\times 2.0}{1.62\times 25.0}=242gmo{l}^{-1}$ Calculated molar mass of benzoic acid $=72+5+12+32+1=122gma{l}^{-1}$ vant Hoff factor, $i=\frac{\text{calculated}\text{molar}\text{mass}}{\text{observed}\text{molar}\text{mass}}$ $=\frac{122}{242}$ $=0.504$ If $\alpha$ is the degree of association of benzoic acid, then, $2C_6{H}_{6}COOH\rightleftharpoons {(C_6{H}_{6}COOH)}_2$ Initial moles 1 0 After association $(1-\alpha )$ $\frac{\alpha }{2}$ Total number of moles of solute after association $=(1-\alpha )+\frac{\alpha }{2}=1-\frac{\alpha }{2}$ $i=\frac{1-\alpha /2}{1}=0.504$ or $1-\frac{\alpha }{2}=0.504$ $\alpha =(1-0.504)\times 2=0.496\times 2$ $=0.992$ or $99.2$