(B) The depression in freezing point $\Delta T_f$ is given by $\Delta T_f = K_f \cdot m$,where $m$ is the molality.Since the mass of solvent $(1 \ kg)

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(B) The depression in freezing point $\Delta T_f$ is given by $\Delta T_f = K_f \cdot m$,where $m$ is the molality.Since the mass of solvent $(1 \ kg)$ and the mass of solute $(1 \ g)$ are the same for both,the molality $m$ is inversely proportional to the molar mass $M$ $(m = \frac{n}{W_{solvent}} = \frac{mass}{M \cdot W_{solvent}})$.Therefore,$\frac{\Delta T_x}{\Delta T_y} = \frac{M_y}{M_x}$.Given $\frac{\Delta T_x}{\Delta T_y} = \frac{1}{4}$,we have $\frac{1}{4} = \frac{M_y}{M_x}$.This implies $\frac{M_x}{M_y} = \frac{4}{1}$,so $M_x : M_y = 4 : 1$.

Class 12 Chemistry Solutions Depression of freezing point of the solvent

Medium

Two solutions $A$ and $B$ are prepared by dissolving $1 \ g$ of non-volatile solutes $X$ and $Y$ respectively in $1 \ kg$ of water. The ratio of depression in freezing points for $A$ and $B$ is found to be $1: 4$. The ratio of molar masses of $X$ and $Y$ is.

JEE MAIN 2022 All JEE MAIN

A
$1: 4$
B
$4: 1$
C
$1: 2$
D
$2: 1$

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Depression of freezing point of the solvent

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Arrange the following solutions in order of decreasing freezing points:
$(a) \ 0.075 \ M \ CuSO_4$ $(b) \ 0.060 \ M \ (NH_4)_2SO_4$
$(c) \ 0.14 \ M \ urea$ $(d) \ 0.04 \ M \ MgCl_2$

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Molal depression constant $(K_{f})$ is dependent on

EasyAP EAMCET 2023

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$A$ solution containing $7.5 \ g$ of urea (molar mass $= 60 \ g \ mol^{-1}$) in $1 \ kg$ of water freezes at the same temperature as another solution containing $15 \ g$ of solute $X$,in the same amount of water. The molar mass of $X \ (g \ mol^{-1})$ is

MediumAP EAMCET 2025

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What is the freezing point depression constant of a solvent,$50 \ g$ of which contains $1 \ g$ of a non-volatile solute (molar mass $256 \ g \ mol^{-1}$) and the decrease in freezing point is $0.40 \ K$?

MediumJEE MAIN 2025

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Column-$I$ (Various solutions) Column-$II$ (Freezing point)

$a$. $0.1 \, M \ BaCl_2$ solution $p$. $271 \, K$

$b$. $0.1 \, M \ NaCl$ solution $q$. $270 \, K$

$c$. $0.1 \, M \ K_3[Fe(CN)_6]$ solution $r$. $268 \, K$

$d$. $0.1 \, M \ Al_2(SO_4)_3$ solution $s$. $269 \, K$

Given: Freezing point of $0.1 \, M$ sucrose solution $= 272 \, K$ and freezing point of water $= 273 \, K$.
Which of the following options shows the correct matches?

Column-$I$ (Various solutions)	Column-$II$ (Freezing point)
$a$. $0.1 \, M \ BaCl_2$ solution	$p$. $271 \, K$
$b$. $0.1 \, M \ NaCl$ solution	$q$. $270 \, K$
$c$. $0.1 \, M \ K_3[Fe(CN)_6]$ solution	$r$. $268 \, K$
$d$. $0.1 \, M \ Al_2(SO_4)_3$ solution	$s$. $269 \, K$

Medium

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Two solutions $A$ and $B$ are prepared by dissolving $1 \ g$ of non-volatile solutes $X$ and $Y$ respectively in $1 \ kg$ of water. The ratio of depression in freezing points for $A$ and $B$ is found to be $1: 4$. The ratio of molar masses of $X$ and $Y$ is.

Similar Questions

Arrange the following solutions in order of decreasing freezing points:$(a) \ 0.075 \ M \ CuSO_4$ $(b) \ 0.060 \ M \ (NH_4)_2SO_4$$(c) \ 0.14 \ M \ urea$ $(d) \ 0.04 \ M \ MgCl_2$

Molal depression constant $(K_{f})$ is dependent on

$A$ solution containing $7.5 \ g$ of urea (molar mass $= 60 \ g \ mol^{-1}$) in $1 \ kg$ of water freezes at the same temperature as another solution containing $15 \ g$ of solute $X$,in the same amount of water. The molar mass of $X \ (g \ mol^{-1})$ is

What is the freezing point depression constant of a solvent,$50 \ g$ of which contains $1 \ g$ of a non-volatile solute (molar mass $256 \ g \ mol^{-1}$) and the decrease in freezing point is $0.40 \ K$?

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Arrange the following solutions in order of decreasing freezing points:
$(a) \ 0.075 \ M \ CuSO_4$ $(b) \ 0.060 \ M \ (NH_4)_2SO_4$
$(c) \ 0.14 \ M \ urea$ $(d) \ 0.04 \ M \ MgCl_2$