When 2 . 44 grams of benzoic acid C 6 H 5 COOH dissolved in 25 grams of benzene, it shows depression of…

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$\begin{array}{cccccc}{W}_B& =& 2.44g , & K_f& =& 5 K kg mo{l}^{-1}\\ W_{A }& =& 25g , & ∆T_{f } & =& 2.2K \end{array}$
$$\begin{gathered} \begin{array}{cccccc}{W}_B& =& 2.44g , & K_f& =& 5 K kg mo{l}^{-1}\\ W_{A }& =& 25g , & ∆T_{f } & =& 2.2K \end{array} \\ Now , \\ ∆T_{f }=\hspace{0.17em}{K}_{f }\times \frac{W_B}{M_{B }}\times \frac{1000}{W_A} \\ {M}_B=\frac{5\times 2.44\times 1000}{2.2\times 25}=221.8g/mol \\ 2C_6{H}_{5}COOH \iff {\left(C_6{H}_{5}COOH \right)}_2 \\ If x is the degree of association , (1-x) mole of \\ benzoic acid left undissociated and corresponding \frac{x}{2} \\ as associated moles of C_6{H}_{5}COOH at equilibrium. \\ Total number of moles at equilibrium \\ 1-x+\frac{x}{2}=1-\frac{x}{2} \\ i=\frac{Normal molecular mass}{Abnormal molecular mass} \\ 1-\frac{x}{2 }=\frac{122}{221.8} \\ x=0.900=90\%\hspace{0.17em} \end{gathered}$$