Answer :

Given: Mass of CH_{3}CH_{2}CHClCOOH = 10 g

Mass of water = 250g

K_{a} = 1.4 × 10^{–3},

K_{f} = 1.86 K kg mol^{–1}

Molar mass of CH_{3}CH_{2}CHClCOOH = 12 + 3 + 12 + 2 + 12 + 1 + 35.5 + 2 + 16 + 16 + 1

= 122.5 g mol^{–1}

Number of moles of solute =

→ No. of moles =

∴ No. of moles = mol

Now, Molality is given as,

M = 0.3264 kg/mol

CH_{3}CH_{2}CHClCOOH CH_{3}CH_{2}CHClCOO^{-} + H ^{+}

Total moles at equilibrium = (1-α) + 2 α

= 1 + α

In order to find out the depression in freezing point, _{f}_{f} values of i(vant Hoff’s factor) and α(degree of dissociation) are to be found out.

To find out degree of dissociation, _{α}

→

Here, the value of α is negligible as compared to 1, and hence 1- = 1, giving,

→

∴

To find out Vant Hoff’s factor,

→

∴

Now, to find out the depression in freezing point,

_{f}_{f}

∴

__Thus, the depression in freezing point is__ _{f} __= 0.65 ^{0}C__

Rate this question :