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}}=2.0\,g $ Mass of solvent (benzene) $ {{w}_{1}}=25.0\,g $ Observed $ \Delta {{T}_{f}}=1.62\,K $ $ {{k}_{f}}=4.9\,K\,kg\,mo{{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}=242\,g\,mo{{l}^{-1}} $ Calculated molar mass of benzoic acid $ =72+5+12+32+1=122\,g\,ma{{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, $ 2{{C}_{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% $