The molal depression constant of a solvent having a freezing point of -23-C and latent heat of fusion of 90 J g-1 is-

Since, Freezing nbsp;point depression constant, K_f = RT_f^21000 × L_f where, nbsp; R = 8.314 J K^ 1 mol^ 1 Freezing point of pure solvent, T_f = 23^∘C = ...

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Standard XII

Chemistry

Depression in Freezing Point

The molal dep...

Question

The molal depression constant of a solvent having a freezing point of $- 23^{\circ} C$ and latent heat of fusion of $90 J g^{- 1}$ is:

3.34 K k g m o l^{- 1}

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5.77 K k g m o l^{- 1}

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0.34 K k g m o l^{- 1}

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8.23 K k g m o l^{- 1}

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Solution

The correct option is B $5.77 K k g m o l^{- 1}$
Since,
Freezing point depression constant,
$K_{f} = \frac{R T_{f}^{2}}{1000 \times L_{f}} where, R = 8.314 J K^{- 1} m o l^{- 1} Freezing point of pure solvent, T_{f} = - 23^{\circ} C = 273 - 23 = 250 K, Latent heat of fusion, L_{f} = 90 J g^{- 1}$
Substituting the values in the above equation,
$K_{f} = \frac{8.314 \times (250)^{2}}{1000 \times 90} = 5.77 K g m o l^{- 1}$

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Depression in Freezing Point

Standard XII Chemistry

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The molal depression constant of a solvent having a freezing point of −23∘C and latent heat of fusion of 90 J g−1 is:

The molal depression constant of a solvent having a freezing point of $- 23^{\circ} C$ and latent heat of fusion of $90 J g^{- 1}$ is: