Kf for water is 1-86-C kg mol-1- What is the molality of a solution which freezes at -0-192-C-

Freezing point depression of a solution is given by, T_f = K_f × m where, T_f is the depression in freezing point. nbsp; K_f is molal depression constant. m i ...

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Byju's Answer

Standard XII

Chemistry

Depression in Freezing Point

Kf for water ...

Question

$K_{f}$ for water is ${1.86}^{\circ} C k g m o l^{- 1} .$ What is the molality of a solution which freezes at $- {0.192}^{\circ} C$ ?

0.103 m

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0.206 m

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0.309 m

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3.09 m

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Solution

The correct option is A $0.103 m$
Freezing point depression of a solution is given by,
$△ T_{f} = K_{f} \times m$
where,
$△ T_{f}$ is the depression in freezing point.
$K_{f}$ is molal depression constant.
m is molality of solution.
Freezing point of the pure water is $0^{\circ}$
Thus, the depression in freezing point of water is ${0.192}^{\circ} C$
$Molality = \frac{△ T_{f}}{K_{f}}$
$Molality = \frac{0.192}{1.86} = 0.103 m$

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Kf for water is 1.86∘C kg mol−1. What is the molality of a solution which freezes at −0.192∘C?

$K_{f}$ for water is ${1.86}^{\circ} C k g m o l^{- 1} .$ What is the molality of a solution which freezes at $- {0.192}^{\circ} C$ ?