(C) The formula for freezing point depression is $\Delta T_{f} = i \times K_{f} \times m$.Here,$\Delta T_{f} = 0.2 \ K$,$K_{f} = 2 \ K \ kg \ mol^{-1}

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(C) The formula for freezing point depression is $\Delta T_{f} = i \times K_{f} \times m$.Here,$\Delta T_{f} = 0.2 \ K$,$K_{f} = 2 \ K \ kg \ mol^{-1}$,and for $NaCl$,the van't Hoff factor $i = 2$.The molality $m = \frac{w / 58.5}{0.6 \ kg}$,where $w$ is the mass of $NaCl$ in grams.Substituting the values: $0.2 = 2 \times 2 \times \frac{w}{58.5 \times 0.6}$.$0.2 = 4 \times \frac{w}{35.1}$.$w = \frac{0.2 \times 35.1}{4} = 1.755 \ g$.

Class 12 Chemistry Solutions Depression of freezing point of the solvent

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How much amount of $NaCl$ should be added to $600 \ g$ of water $(\rho=1.00 \ g / mL)$ to decrease the freezing point of water to $-0.2^{\circ} C ?$ ............. $gm$
(The freezing point depression constant for water $=2 \ K \ kg \ mol^{-1}$ )

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A
$2.25$
B
$2$
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$1.75$
D
$1.5$

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How much amount of $NaCl$ should be added to $600 \ g$ of water $(\rho=1.00 \ g / mL)$ to decrease the freezing point of water to $-0.2^{\circ} C ?$ ............. $gm$(The freezing point depression constant for water $=2 \ K \ kg \ mol^{-1}$ )

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How much amount of $NaCl$ should be added to $600 \ g$ of water $(\rho=1.00 \ g / mL)$ to decrease the freezing point of water to $-0.2^{\circ} C ?$ ............. $gm$
(The freezing point depression constant for water $=2 \ K \ kg \ mol^{-1}$ )