(A) Given:$K_f = 1.86 \ K \ kg \ mol^{-1}$Mass of solvent $(W_A)$ $= 4 \ kg = 4000 \ g$Freezing point depression $\Delta T_f = 0 - (-6) = 6 \ K$Molar

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(A) Given:$K_f = 1.86 \ K \ kg \ mol^{-1}$Mass of solvent $(W_A)$ $= 4 \ kg = 4000 \ g$Freezing point depression $\Delta T_f = 0 - (-6) = 6 \ K$Molar mass of solute $(M_B)$ $= 62 \ g \ mol^{-1}$The formula for freezing point depression is $\Delta T_f = K_f \times m$,where $m$ is molality.$\Delta T_f = K_f \times \frac{w \times 1000}{M_B \times W_A \text{ (in g)}}$$6 = 1.86 \times \frac{w \times 1000}{62 \times 4000}$$6 = 1.86 \times \frac{w}{62 \times 4}$$w = \frac{6 \times 62 \times 4}{1.86}$$w = \frac{1488}{1.86} = 800 \ g$

Class 12 Chemistry Solutions Depression of freezing point of the solvent

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Ethylene glycol is used as an antifreeze in cold climates. The mass of ethylene glycol $(C_2H_6O_2)$ that should be added to $4 \ kg$ of water to prevent it from freezing at $-6 \ ^oC$ is ......... $g$.
($K_f$ for water $= 1.86 \ K \ kg \ mol^{-1}$,and molar mass of ethylene glycol $= 62 \ g \ mol^{-1}$)

AIEEE 2011 All AIEEE

A
$800.00$
B
$204.30$
C
$400.00$
D
$304.60$

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Depression of freezing point of the solvent

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Freezing point of a $4 \%$ aqueous solution of $X$ is equal to the freezing point of a $12 \%$ aqueous solution of $Y$. If the molecular weight of $X$ is $A$,then the molecular weight of $Y$ is ............. $A$.

DifficultJEE MAIN 2019

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Statement $1$: At the freezing point,the solid substance crystallizes from the solution.
Statement $2$: Depression of freezing point is the difference between the freezing point of the solvent and the freezing point of the solution.

Medium

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Calculate the mass of ascorbic acid $(\text{C}_6\text{H}_8\text{O}_6)$ to be dissolved in $75 \text{ g}$ of acetic acid to lower its melting point by $1.5^\circ\text{C}$. $[K_f = 3.9 \text{ K kg mol}^{-1}, \text{H} = 1, \text{C} = 12, \text{O} = 16 \text{ amu}]$ (in $\text{ g}$)

DifficultGUJCET 2026

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$5 \ g$ sucrose (molar mass $= 342 \ g \ mol^{-1}$) is dissolved in $100 \ g$ of solvent,decreasing the freezing point by $2.15 \ K$. What is the cryoscopic constant $(K_{f})$ of the solvent?

MediumMHT CET 2021

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

EasyMHT CET 2022

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Freezing point of a $4 \%$ aqueous solution of $X$ is equal to the freezing point of a $12 \%$ aqueous solution of $Y$. If the molecular weight of $X$ is $A$,then the molecular weight of $Y$ is ............. $A$.

Statement $1$: At the freezing point,the solid substance crystallizes from the solution.Statement $2$: Depression of freezing point is the difference between the freezing point of the solvent and the freezing point of the solution.

Calculate the mass of ascorbic acid $(\text{C}_6\text{H}_8\text{O}_6)$ to be dissolved in $75 \text{ g}$ of acetic acid to lower its melting point by $1.5^\circ\text{C}$. $[K_f = 3.9 \text{ K kg mol}^{-1}, \text{H} = 1, \text{C} = 12, \text{O} = 16 \text{ amu}]$ (in $\text{ g}$)

$5 \ g$ sucrose (molar mass $= 342 \ g \ mol^{-1}$) is dissolved in $100 \ g$ of solvent,decreasing the freezing point by $2.15 \ K$. What is the cryoscopic constant $(K_{f})$ of the solvent?

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

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Statement $1$: At the freezing point,the solid substance crystallizes from the solution.
Statement $2$: Depression of freezing point is the difference between the freezing point of the solvent and the freezing point of the solution.