Find the depression in freezing point of a so | vedclass

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

Question

(B) The formula for the depression in freezing point is $\Delta T_{f} = K_{f} 	imes m$,where $m$ is the molality of the solution.Molality $m = \frac{

Vedclass · Accepted Answer

$1.5$

Vedclass · Answer

(B) The formula for the depression in freezing point is $\Delta T_{f} = K_{f} 	imes m$,where $m$ is the molality of the solution.Molality $m = \frac{W_2 	imes 1000}{M_2 	imes W_1}$,where $W_2$ is the mass of the solute,$M_2$ is the molar mass of the solute,and $W_1$ is the mass of the solvent in grams.Substituting the given values: $W_2 = 3.2 \ g$,$M_2 = 128 \ g \ mol^{-1}$,$W_1 = 80 \ g$,and $K_{f} = 4.8 \ K \ kg \ mol^{-1}$.$\Delta T_{f} = \frac{4.8 \ K \ kg \ mol^{-1} 	imes 3.2 \ g 	imes 1000 \ g \ kg^{-1}}{128 \ g \ mol^{-1} 	imes 80 \ g}$.$\Delta T_{f} = \frac{15360}{10240} \ K = 1.5 \ K$.

Find the depression in freezing point of a solution when $3.2 \ g$ of a non-volatile solute with a molar mass of $128 \ g \ mol^{-1}$ is dissolved in $80 \ g$ of solvent,given that the cryoscopic constant of the solvent is $4.8 \ K \ kg \ mol^{-1}$. (in $K$)

Similar Questions

Calculate the molality of a nonvolatile solution if the solution freezes at $-0.95^{\circ}C$ $[K_f \text{ for water} = 1.86 \ K \ kg \ mol^{-1}, \text{ freezing point of water} = 0^{\circ}C]$.

What happens to the freezing point of benzene when a small quantity of naphthalene is added to it?

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.

When a solution containing a non-volatile solute freezes,which equilibrium exists?

If the $K_f$ value of $H_2O$ is $1.86$,what is the value of $\Delta T_f$ for a $0.1 \ m$ solution of a non-volatile solute?

Vedclass Test Series

Exam Paper Generator

Online Exam Module