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Standard XII

Chemistry

Depression in Freezing Point

The depressio...

Question

The depression in freezing point is $0.24$ K obtained by dissolving $1$ gram kel in $200$ gram water, calculate Van't Hoff factor. The molal depression constant is $1.86$ K kg $m o l e^{- 1}$ .

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Solution

$Δ T_{f} = 0.24 k$ mass of $K C l = 1 g$ $K f = 1.86 K k g / m o l$
mass of water (solvent) $= 200 g$
Molar mass of solute $= \frac{1000 \times K f \times mass of solute}{Δ T_{1} \times mass of solvent}$
$= \frac{1000 \times 1.86 \times 1}{0.24 \times 200} = 38.75 g / m o l$
This is the experimental molar mass of solute $(K C l)$ .
Theoretical molar mass $= 3 a + 35.5 = 74.5 g / m o l$
van't hott factor (i) $= \frac{Theoretical molar mass}{Expermenmtal molar mass} = \frac{74.5}{38.75}$
$∴ i = 1.92$

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The depression in freezing point is 0.24K obtained by dissolving 1 gram kel in 200 gram water, calculate Van't Hoff factor. The molal depression constant is 1.86 K kg mole−1.

The depression in freezing point is $0.24$ K obtained by dissolving $1$ gram kel in $200$ gram water, calculate Van't Hoff factor. The molal depression constant is $1.86$ K kg $m o l e^{- 1}$ .