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(D) The boiling point of the solution is $T_b = T_b^o + \Delta T_b = T_b^o + i K_b m$.The freezing point of the solution is $T_f = T_f^o - \Delta T_f

Vedclass · Accepted Answer

$0.77$

Vedclass · Answer

(D) The boiling point of the solution is $T_b = T_b^o + \Delta T_b = T_b^o + i K_b m$.The freezing point of the solution is $T_f = T_f^o - \Delta T_f = T_f^o - i K_f m$.The difference is given as $T_b - T_f = 75.7 \ ^\circ C$.Substituting the expressions: $(T_b^o + i K_b m) - (T_f^o - i K_f m) = 75.7$.$(80 + i 	imes 2.65 	imes 0.2) - (5.5 - i 	imes 5.12 	imes 0.2) = 75.7$.$(80 - 5.5) + i 	imes 0.2 	imes (2.65 + 5.12) = 75.7$.$74.5 + i 	imes 0.2 	imes (7.77) = 75.7$.$i 	imes 1.554 = 75.7 - 74.5 = 1.2$.$i = \frac{1.2}{1.554} \approx 0.77$.

The difference between the boiling point and the freezing point of a $0.2 \ m$ solution of acetic acid in benzene is $75.7 \ ^\circ C$. Calculate the value of the van't Hoff factor $i$. (For benzene,$K_b = 2.65 \ K \ m^{-1}$,$K_f = 5.12 \ K \ m^{-1}$,$T_b^o = 80 \ ^\circ C$,$T_f^o = 5.5 \ ^\circ C$)

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