A) 78.2%
B) 82.6%
C) 89.7%
D) 99.2%
Correct Answer: D
Solution :
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 \[99.2%\]
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