(B) The depression in freezing point is given by $\Delta T_f = i \times K_f \times m$. Since $K_f$ and $m$ are constant,$\Delta T_f \propto i$. Freezi

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(B) The depression in freezing point is given by $\Delta T_f = i \times K_f \times m$. Since $K_f$ and $m$ are constant,$\Delta T_f \propto i$. Freezing point $T_f = T_f^0 - \Delta T_f = 273 - (i \times K_f \times 0.1)$. For sucrose,$i=1$,$\Delta T_f = 273 - 272 = 1 \, K$. Thus,$K_f \times 0.1 = 1$. $T_f = 273 - i$. For $a. \ BaCl_2$ $(i=3)$: $T_f = 273 - 3 = 270 \, K$ $(q)$. For $b. \ NaCl$ $(i=2)$: $T_f = 273 - 2 = 271 \, K$ $(p)$. For $c. \ K_3[Fe(CN)_6]$ $(i=4)$: $T_f = 273 - 4 = 269 \, K$ $(s)$. For $d. \ Al_2(SO_4)_3$ $(i=5)$: $T_f = 273 - 5 = 268 \, K$ $(r)$. Therefore,the correct match is $a-q, b-p, c-s, d-r$.

Class 12 Chemistry Solutions Depression of freezing point of the solvent

Medium

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	$s$. $269 \, K$

A
$a-r, b-p, c-s, d-q$
B
$a-q, b-p, c-s, d-r$
C
$a-q, b-s, c-p, d-r$
D
$a-s, b-q, c-r, d-q$

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Depression of freezing point of the solvent

$1.00 \text{ g}$ of a non-electrolyte solute dissolved in $50 \text{ g}$ of benzene lowered the freezing point of benzene by $0.40 \text{ K}$. The freezing point depression constant of benzene is $5.12 \text{ K kg mol}^{-1}$. Find the molar mass of the solute.

EasyGSEB 2021

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$K_f$ for water is $1.86 \, K \, kg \, mol^{-1}$. If your automobile radiator holds $1.0 \, kg$ of water,how many $gm$ of ethylene glycol $(C_2H_6O_2)$ must you add to lower the freezing point of the solution to $-2.8 \, ^oC$?

DifficultAIEEE 2012

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At $T$ $(K)$,$x \ g$ of a non-volatile solid (molar mass $78 \ g \ mol^{-1}$) when added to $0.5 \ kg$ water,lowered its freezing point by $1.0^{\circ} C$. What is $x$ (in $g$)? ($K_{f}$ of water at $T$ $(K)$ = $1.86 \ K \ kg \ mol^{-1}$)

DifficultAP EAMCET 2022

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Calculate the molality of a solution of a non-volatile solute having a depression in freezing point of $0.93 \ K$ and a cryoscopic constant of the solvent of $1.86 \ K \ kg \ mol^{-1}$.

MediumMHT CET 2024

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The molal depression constant for water is $1.86 \, ^\circ C \, kg \, mol^{-1}$. The freezing point of a $0.05 \, m$ solution of a non-electrolyte in water is:

Medium

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$1.00 \text{ g}$ of a non-electrolyte solute dissolved in $50 \text{ g}$ of benzene lowered the freezing point of benzene by $0.40 \text{ K}$. The freezing point depression constant of benzene is $5.12 \text{ K kg mol}^{-1}$. Find the molar mass of the solute.

$K_f$ for water is $1.86 \, K \, kg \, mol^{-1}$. If your automobile radiator holds $1.0 \, kg$ of water,how many $gm$ of ethylene glycol $(C_2H_6O_2)$ must you add to lower the freezing point of the solution to $-2.8 \, ^oC$?

At $T$ $(K)$,$x \ g$ of a non-volatile solid (molar mass $78 \ g \ mol^{-1}$) when added to $0.5 \ kg$ water,lowered its freezing point by $1.0^{\circ} C$. What is $x$ (in $g$)? ($K_{f}$ of water at $T$ $(K)$ = $1.86 \ K \ kg \ mol^{-1}$)

Calculate the molality of a solution of a non-volatile solute having a depression in freezing point of $0.93 \ K$ and a cryoscopic constant of the solvent of $1.86 \ K \ kg \ mol^{-1}$.

The molal depression constant for water is $1.86 \, ^\circ C \, kg \, mol^{-1}$. The freezing point of a $0.05 \, m$ solution of a non-electrolyte in water is:

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