Question

14)
The molal depression constant of a solvent whose
freezing point is -23°C and latent heat of fusion is
125 Jg 'will be
(1) 4.16 K kg mol-1.
(2) 2.25 K kg mol-1
(3) 3.12 K kg mot
(4) 1.75 K kg mol-1
​

Answer 1

(1) 4.16 K kg mol⁻¹ is the molal depression constant of a solvent.

Explanation:

Given the Freezing point is -23°C that is 250K

Latent heat of fusion is 125Jg⁻¹

The molal depression constant is given by $K=\frac{MRT^{2} }{Hfusion} =\frac{RT^{2} }{Lfusion}$....(1)

but Latent heat of fusion is given per gram and options are in Kg

⇒Latent heat of fusion of solvent is $Lfusion=$125*10³ J Kg⁻¹

substituting all the given values in equation (1)

we get, $K=\frac{8.314*250^{2} }{125*10*10*10}$=$4.157$ K kg mol⁻¹

therefore, molal depression constant of the solvent is $K=4.16 \frac{K kg}{mol}$ .