An aqueous solution of a weak monobasic acid | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 $\text{g/mol}$?

Similar Questions

When an aqueous solution of glucose is frozen,which crystals will separate out first?

In winter,the normal temperature in Kullu valley was found to be $-11\,^{\circ}C$. Is a $28\%$ (by mass) aqueous solution of ethylene glycol suitable for a car radiator? $K_f$ for water $= 1.86\, K\, kg\, mol^{-1}$.

The freezing point of an aqueous solution containing $25 \ g$ of ethanol $(C_2H_5OH)$ in $1000 \ g$ of $H_2O$ is $(K_f = 1.86 \ K \ kg \ mol^{-1})$ (in $^{\circ} C$)

Camphor is used in the determination of molecular mass because $....$

$A$ $5 \%$ solution (by mass) of cane sugar in water has a freezing point of $271 \, K$. Calculate the freezing point of a $5 \%$ glucose solution in water,given that the freezing point of pure water is $273.15 \, K$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module