Three particles of a solute A associate in be | vedclass

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(A) For association,the van't Hoff factor $i$ is given by $i = 1 + (\frac{1}{n} - 1) \alpha$.Given $n = 3$ and $\alpha = 0.80$,we have $i = 1 + (\frac

Vedclass · Accepted Answer

$278.05$

Vedclass · Answer

(A) For association,the van't Hoff factor $i$ is given by $i = 1 + (\frac{1}{n} - 1) \alpha$.Given $n = 3$ and $\alpha = 0.80$,we have $i = 1 + (\frac{1}{3} - 1) 	imes 0.80 = 1 - 0.5333 = 0.4667$.Depression in freezing point is $\Delta T_f = i \cdot K_f \cdot m$.$\Delta T_f = 0.4667 	imes 5.12 	imes 0.25 = 0.597 \ K \approx 0.6 \ K$.The freezing point of the solution $T_s = T_f^o - \Delta T_f$.$T_f^o$ of benzene in Kelvin is $273.15 + 5.5 = 278.65 \ K$.$T_s = 278.65 - 0.6 = 278.05 \ K$.

Three particles of a solute $A$ associate in benzene to form a trimer $A_3$. Calculate the freezing point of a $0.25 \ m$ solution if the degree of association of solute $A$ is $0.80$. The freezing point of benzene is $5.5 \ ^oC$ and $K_f$ is $5.12 \ K \ kg \ mol^{-1}$. (Answer in $K$)

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