Question

When HNO2 is dissolved in water it partially dissociates according to the equation HNO2⇌H++NO−2. A solution is prepared that contains 4.000 g of HNO2 in 1.000 kg of water. Its freezing point is found to be -0.1692 ∘C Calculate the fraction of HNO2 that has dissociated.

Answer 1

Answer:

The degree of dissociation is 1

thus the fraction dissociated is one

Explanation:

The depression in freezing point is a colligative property and depends upon the number of solute particles present in the solution.

The relation between depression in freezing point and molality is:

$depressioninfreezingpoint=(i)K_{f}Xmolality$

For water

Kf=1.86 °C/m

Where

i= Van't Hoff factor

For electrolytes the i depends upon on the extent of dissociation

i = αn + (1 - α)

Where

α = is degree of dissociation

Let us put the values

$moles=\frac{mass}{molarmass}=\frac{4}{47}=0.085$

$0.1692=i\frac{0.0851}{1}\\ i=2$

Putting value:

$2 = \alpha n + (1 - \alpha )\\n=2\\\alpha =1$