(C) The formula for depression in freezing point is $\Delta T_f = K_f \times m$. Here,$\Delta T_f = 0.2 \ K$,$K_f = 1.71 \ K \ kg \ mol^{-1}$,$M_B = 6

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(C) The formula for depression in freezing point is $\Delta T_f = K_f \times m$. Here,$\Delta T_f = 0.2 \ K$,$K_f = 1.71 \ K \ kg \ mol^{-1}$,$M_B = 60 \ g \ mol^{-1}$,and $W_A = 98 \ g$. The molality $m$ is given by $\frac{W_B \times 1000}{M_B \times W_A}$. Substituting the values: $0.2 = 1.71 \times \frac{W_B \times 1000}{60 \times 98}$. Solving for $W_B$: $W_B = \frac{0.2 \times 60 \times 98}{1.71 \times 1000} = \frac{1176}{1710} \approx 0.687 \ g$.

Class 12 Chemistry Solutions Depression of freezing point of the solvent

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What is the mass of solute having molar mass $60 \ g \ mol^{-1}$ when dissolved in $98 \ g$ of solvent decreases its freezing point by $0.2 \ K$ (in $g$)? (The numerical value of cryoscopic constant of solvent is $1.71 \ K \ kg \ mol^{-1}$)

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A
$0.5$
B
$1.5$
C
$0.687$
D
$2.0$

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Depression of freezing point of the solvent

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The freezing point depression constant $(K_{f})$ of benzene is $5.12 \ K \ kg \ mol^{-1}$. The freezing point depression for the solution of molality $0.078 \ m$ containing a non-electrolyte solute in benzene is..........$K$ (rounded off up to two decimal places).

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The freezing point depression of a solution containing $0.6 \ g$ of urea (molar mass $= 60 \ g \ mol^{-1}$) in $100 \ mL$ of benzene (in $K$) is ($K_f$ of benzene $= 4.0 \ K \ kg \ mol^{-1}$).

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Camphor is used in the determination of molecular mass because $....$

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The molar freezing point constant for water is $1.86\,^{\circ}C\,kg\,mol^{-1}$. If $342\,g$ of cane sugar $(C_{12}H_{22}O_{11})$ are dissolved in $1000\,g$ of water,the solution will freeze at $............\,^{\circ}C$.

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How many grams of glucose must be added to $1 \ L$ of water such that the value of $\frac{\Delta T_f}{K_f}$ becomes $\frac{1}{1000}$?

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What is the mass of solute having molar mass $60 \ g \ mol^{-1}$ when dissolved in $98 \ g$ of solvent decreases its freezing point by $0.2 \ K$ (in $g$)? (The numerical value of cryoscopic constant of solvent is $1.71 \ K \ kg \ mol^{-1}$)

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The freezing point depression constant $(K_{f})$ of benzene is $5.12 \ K \ kg \ mol^{-1}$. The freezing point depression for the solution of molality $0.078 \ m$ containing a non-electrolyte solute in benzene is..........$K$ (rounded off up to two decimal places).

The freezing point depression of a solution containing $0.6 \ g$ of urea (molar mass $= 60 \ g \ mol^{-1}$) in $100 \ mL$ of benzene (in $K$) is ($K_f$ of benzene $= 4.0 \ K \ kg \ mol^{-1}$).

Camphor is used in the determination of molecular mass because $....$

The molar freezing point constant for water is $1.86\,^{\circ}C\,kg\,mol^{-1}$. If $342\,g$ of cane sugar $(C_{12}H_{22}O_{11})$ are dissolved in $1000\,g$ of water,the solution will freeze at $............\,^{\circ}C$.

How many grams of glucose must be added to $1 \ L$ of water such that the value of $\frac{\Delta T_f}{K_f}$ becomes $\frac{1}{1000}$?

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