(D) Given: Mass of solute $(W_1)$ = $4.5 \ g$ Mass of solvent $(W_2)$ = $100 \ g$ Depression in freezing point $(\Delta T_f)$ = $0.465 \ K$ Cryoscopic

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(D) Given: Mass of solute $(W_1)$ = $4.5 \ g$ Mass of solvent $(W_2)$ = $100 \ g$ Depression in freezing point $(\Delta T_f)$ = $0.465 \ K$ Cryoscopic constant $(K_f)$ = $1.86 \ K \ kg \ mol^{-1}$ The formula for molar mass $(M_1)$ is: $M_1 = \frac{K_f \times 1000 \times W_1}{\Delta T_f \times W_2}$ Substituting the values: $M_1 = \frac{1.86 \times 1000 \times 4.5}{0.465 \times 100}$ $M_1 = \frac{1.86 \times 45}{0.465}$ $M_1 = 180 \ g/mol$

Class 12 Chemistry Solutions Depression of freezing point of the solvent

Medium

When $4.5 \ g$ of a non-electrolyte solute is dissolved in $100 \ g$ of water,the freezing point of the solution is lowered by $0.465^o C$. The molar mass of the solute is ....... $g/mol$. (Given $K_f = 1.86 \ K \ kg \ mol^{-1}$)

A
$135$
B
$172$
C
$90$
D
$180$

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Depression of freezing point of the solvent

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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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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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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Which of the following aqueous solutions has the highest freezing point?

MediumKCET 2015

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$A$ 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.

Medium

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When $4.5 \ g$ of a non-electrolyte solute is dissolved in $100 \ g$ of water,the freezing point of the solution is lowered by $0.465^o C$. The molar mass of the solute is ....... $g/mol$. (Given $K_f = 1.86 \ K \ kg \ mol^{-1}$)

Similar Questions

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.

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 $T$ $(K)$ = $1.86 \ K \ kg \ mol^{-1}$)

Which of the following aqueous solutions has the highest freezing point?

$A$ 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.

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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.