The molar freezing point constant for water i | vedclass

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

Question

(A) The molar mass of cane sugar $(C_{12}H_{22}O_{11})$ is $342\,g\,mol^{-1}$.Given mass of solute $(w_2)$ = $342\,g$,mass of solvent $(w_1)$ = $1000\

Vedclass · Accepted Answer

$ - 1.86$

Vedclass · Answer

(A) The molar mass of cane sugar $(C_{12}H_{22}O_{11})$ is $342\,g\,mol^{-1}$.Given mass of solute $(w_2)$ = $342\,g$,mass of solvent $(w_1)$ = $1000\,g$,and $K_f = 1.86\,^{\circ}C\,kg\,mol^{-1}$.Molality $(m)$ = $\frac{w_2 	imes 1000}{M_2 	imes w_1} = \frac{342 	imes 1000}{342 	imes 1000} = 1\,mol\,kg^{-1}$.Depression in freezing point $(\Delta T_f)$ = $K_f 	imes m = 1.86 	imes 1 = 1.86\,^{\circ}C$.Freezing point of solution $(T_f)$ = $T_f^{\circ} - \Delta T_f = 0\,^{\circ}C - 1.86\,^{\circ}C = - 1.86\,^{\circ}C$.

The molar freezing point constant for water is $1.86\,^{\circ}C\,kg\,mol^{-1}$. If $342\,g$ of cane sugar $(C_{12}H_{22}O_{11})$ are dissolved in $1000\,g$ of water,the solution will freeze at $............\,^{\circ}C$.

Similar Questions

Calculate the molar mass of a non-volatile solute when $1 \ g$ of it is dissolved in $100 \ g$ of solvent,which decreases its freezing point by $0.2 \ K$. Given: $K_{f} = 1.2 \ K \ kg \ mol^{-1}$.

Calculate the molality of a solution of a non-volatile solute having a depression in freezing point of $0.93 \ K$ and a cryoscopic constant of the solvent of $1.86 \ K \ kg \ mol^{-1}$.

Calculate the cryoscopic constant $(K_f)$ of a solvent if the depression in freezing point of a $0.18 \ m$ solution of a non-volatile solute is $0.2 \ K$.

The depression in freezing point of water observed for the same amount of acetic acid,trichloroacetic acid,and trifluoroacetic acid increases in the order given above. Explain briefly.

If the freezing point of an aqueous urea solution is $271.14 \ K$ at $1 \ \text{atm}$ pressure (given $K_f$ of water = $1.86 \ K \ kg/mol$),then what is the mole fraction of urea in this solution? (Freezing point of pure water is $273 \ K$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module