question_answer

2.0 g of benzoic acid dissolved in 25.0 g benzene shows a depression in freezing poi equal to 1.62 K. Molal depression constant \[\text{Vc}=\text{1}.\text{5V}\] benzene is\[\text{1}00\text{ }\mu \text{A}\]. The percenta association of the acid is

A) 80%

B)                 99%

C) 75%                                       

D) 100%                 \[1.5\mu \] \[\mu \] Normal molecular weight of \[W\] \[\frac{4W}{3}\]=122 \[\frac{5W}{2}\] \[\frac{\pi }{2}\] \[\sigma =\text{5}.\text{67}\times \text{1}{{0}^{-\text{8}}}\text{W}-{{\text{m}}^{\text{2}}}{{\text{K}}^{\text{-4}}}\] \[2{{C}_{6}}{{H}_{5}}COOH\xrightarrow{{}}{{({{C}_{6}}{{H}_{5}}COOH)}_{2}}\] \[t\]                       where, \[{{(Kg)}^{1/2}}\] \[{{(Kg)}^{-1/2}}\] \[{{(Kg)}^{2}}\] % association \[{{(Kg)}^{-2}}\] \[\frac{pV}{nT}\]