Question

the lowering in freezing point of 0.75 molal aqueous solution of nacl (80%of dissociation) is​

Answer 1

To find:

The lowering in freezing point of 0.75 molal aqueous solution of NaCl (80%of dissociation)?

Calculation:

First, let's find the Van't Hoff Factor:

$NaCl \rightarrow Na^{+}+Cl^{-}$

$1 \: \: \: \: \: \: \: \: \: \: \: \: \: \: 0 \: \: \: \: \: \: \: \: \: \: \: \: 0 \: \: \: . \: . \: .(at \: t = 0)$

$1 - x \: \ \: \: \: \: \: \: \: x\: \: \: \: \: \: \: \: \: \: \: \: x \: \: \: . \: . \: .(at \: t =t)$

Now, Van't Hoff Factor be 'i':

$i = \dfrac{x + x + (1 - x)}{1}$

$\implies i = \dfrac{1 + x}{1}$

• Now, x = 80% = 0.8

$\implies i = \dfrac{1 + 0.8}{1}$

$\implies i = 1.8$

Now, depression in freezing point is:

$\Delta T = i \times k_{f} \times m$

$\implies\Delta T = 1.8 \times 1.86\times 0.75$

$\implies\Delta T = 2.5 \: K$

So, depression in freezing point is 2.5 K.