1.00 g of a non-electrolyte solute (molar ma | 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

$0.4$

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: $W_1 = 1.00 \ g$,$M_1 = 250 \ g \ mol^{-1}$,$W_2 = 51.2 \ g$,$K_f = 5.12 \ K \ kg \ mol^{-1}$.Substituting the values: $\Delta T_f = 5.12 	imes \frac{1.00 	imes 1000}{250 	imes 51.2}$.$\Delta T_f = 5.12 	imes \frac{1000}{12800} = 5.12 	imes 0.078125 = 0.4 \ K$.

$1.00 \ g$ of a non-electrolyte solute (molar mass $250 \ g \ mol^{-1}$) was dissolved in $51.2 \ g$ of benzene. If the freezing point depression constant,$K_f$ of benzene is $5.12 \ K \ kg \ mol^{-1},$ the freezing point of benzene will be lowered by .......... $K$.

Similar Questions

$A$ fixed amount of glucose is dissolved in $100 \text{ g}$ water to form a solution that freezes at $-0.2^\circ\text{C}$. If the solution is cooled down to $-0.25^\circ\text{C}$, then ......... $\text{g}$ of ice would have separated.

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

The freezing point of a $0.01 \ m$ aqueous glucose solution is $-0.18^\circ C$. If an equal volume of $0.002 \ m$ glucose solution is added to it,the freezing point of the resulting solution will be ...... $^\circ C$.

$A$ solution of $5 \ g$ nonvolatile solute in $50 \ g$ water decreases its freezing point by $0.2 \ K$. Calculate the molar mass of solute if $K_{f}$ of water is $1.86 \ K \ kg \ mol^{-1}$.

Given that $\Delta T_f$ is the depression in freezing point of the solvent in a solution of a non-volatile solute of molality $m$,the quantity $\lim_{m \to 0} \left( \frac{\Delta T_f}{m} \right)$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module