An aqueous solution of a weak monobasic acid | vedclass

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(B) The depression in freezing point is $\Delta T_f = T_f^\circ - T_f$. Taking the freezing point of pure water $T_f^\circ = 273 	ext{ K}$,we have: $

Vedclass · Accepted Answer

$46$

Vedclass · Answer

(B) The depression in freezing point is $\Delta T_f = T_f^\circ - T_f$. Taking the freezing point of pure water $T_f^\circ = 273 	ext{ K}$,we have: $\Delta T_f = 273 - 272.813 = 0.187 	ext{ K}$. The molecular mass $(M)$ is calculated using the formula: $M = \frac{K_f 	imes w 	imes 1000}{\Delta T_f 	imes W}$ Substituting the given values: $M = \frac{1.86 	imes 0.1 	imes 1000}{0.187 	imes 21.7} = \frac{186}{4.0579} \approx 45.83 	ext{ g/mol}$. The molecular mass is approximately $46 	ext{ g/mol}$.

An aqueous solution of a weak monobasic acid containing $0.1 \text{ g}$ in $21.7 \text{ g}$ of water freezes at $272.813 \text{ K}$. If the value of $K_f$ for water is $1.86 \text{ K kg/mol}$,what is the molecular mass of the monobasic acid in $\text{g/mol}$?

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