2 g of a non-electrolyte solute (molar mass | vedclass

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

Question

(B) The formula for freezing point depression is $\Delta T_f = K_f 	imes m$,where $m$ is the molality of the solution. Molality $m = \frac{w_1 	imes

Vedclass · Accepted Answer

$0.3$

Vedclass · Answer

(B) The formula for freezing point depression is $\Delta T_f = K_f 	imes m$,where $m$ is the molality of the solution. Molality $m = \frac{w_1 	imes 1000}{m_1 	imes w_2}$,where $w_1$ is the mass of solute,$m_1$ is the molar mass of solute,and $w_2$ is the mass of solvent in grams. Given: $w_1 = 2 \ g$,$m_1 = 500 \ g \ mol^{-1}$,$w_2 = 57.3 \ g$,$K_f = 4.3 \ K \ kg \ mol^{-1}$. Substituting the values: $\Delta T_f = 4.3 	imes \frac{2 	imes 1000}{500 	imes 57.3}$. $\Delta T_f = 4.3 	imes \frac{2000}{28650} = 4.3 	imes 0.0698 \approx 0.3 \ K$.

$2 \ g$ of a non-electrolyte solute (molar mass is $500 \ g \ mol^{-1}$) was dissolved in $57.3 \ g$ of xylene. If the freezing point depression constant $K_f$ of xylene is $4.3 \ K \ kg \ mol^{-1}$,then the depression in freezing point of xylene is.......... (in $K$)

Similar Questions

Calculate the freezing point of a $0.05 \ m$ aqueous solution of a non-electrolyte. (in $K$)

During the depression of freezing point in a solution,which of the following are in equilibrium?

What is the unit of cryoscopic constant?

When glycerine is added to a litre of water,which of the following behaviors is observed?

When $0.01 \ mol$ of sugar is dissolved in $100 \ g$ of a solvent,the depression in freezing point is $0.40 \ ^oC$. When $0.03 \ mol$ of glucose is dissolved in $50 \ g$ of the same solvent,the depression in freezing point will be $......... \ ^oC$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module