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(B) For a weak monobasic acid $HA$,the dissociation is $HA \rightleftharpoons H^+ + A^-$. The van't Hoff factor $i = 1 + \alpha$,where $\alpha = 0.3$.

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$30$

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(B) For a weak monobasic acid $HA$,the dissociation is $HA ightleftharpoons H^+ + A^-$. The van't Hoff factor $i = 1 + \alpha$,where $\alpha = 0.3$. So,$i = 1 + 0.3 = 1.3$. The observed freezing point depression is $(\Delta T_f)_{obs} = i 	imes K_f 	imes m$. The theoretical freezing point depression is $(\Delta T_f)_{cal} = K_f 	imes m$. The percentage increase in freezing point depression is given by $\frac{(\Delta T_f)_{obs} - (\Delta T_f)_{cal}}{(\Delta T_f)_{cal}} 	imes 100$. Substituting the values: $\frac{i 	imes K_f 	imes m - K_f 	imes m}{K_f 	imes m} 	imes 100 = (i - 1) 	imes 100$. Percentage increase $= (1.3 - 1) 	imes 100 = 0.3 	imes 100 = 30\%$.

If the degree of dissociation of an aqueous solution of a weak monobasic acid is determined to be $0.3$,then the observed freezing point depression will be $.....\%$ higher than the expected/theoretical freezing point depression. (Nearest integer)

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