An aqueous solution is prepared by dissolving 0.1 ~mol of an ionic salt in 1.8 ~kg of water at 35^\circ C . The salt remains 90 \% dissociated in the solution. The vapour pressure of the solution is 59.724 ~mm of Hg. Vapor pressure of water at 35 ^\circ C is 60.000 ~mm of Hg . The number of ions present per formula unit of the ionic salt is \underline\hspace2cm .

Solutions PYQ — JEE Advanced 2022 · Shift 2 · Q2 with Solution

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Correct answer: 5

For a non-volatile solute,

\frac{p^0-p}{p^0}=x_{\text{solute particles}}

where $x_{\text{solute particles}}$ is the mole fraction of total solute particles actually present in solution.

Given:

p^0=60.000\ \text{mm Hg},\qquad p=59.724\ \text{mm Hg}

So,

\frac{p^0-p}{p^0}=\frac{60.000-59.724}{60.000}=\frac{0.276}{60.000}=0.0046

Water taken = $1.8\ \text{kg}=1800\ \text{g}$

n_{\text{water}}=\frac{1800}{18}=100\ \text{mol}

Degree of dissociation:

\alpha=0.9

For an ionic solute, van't Hoff factor is

i=1+\alpha(\nu-1)

If initial moles of salt = $0.1$ , then effective moles of solute particles are

0.1\,i

Hence mole fraction of solute particles is

x_{\text{solute particles}}=\frac{0.1i}{100+0.1i}

Given this equals $0.0046$ :

\frac{0.1i}{100+0.1i}=0.0046

0.1i=0.0046(100+0.1i)

0.1i=0.46+0.00046i

0.1i-0.00046i=0.46

0.09954i=0.46

i\approx \frac{0.46}{0.09954}\approx 4.62

Since this is very close to $4.6$ , we take

i=4.6

i=1+0.9(\nu-1)

Substitute $i=4.6$ :

4.6=1+0.9(\nu-1)

3.6=0.9(\nu-1)

\nu-1=4

\nu=5

The number of ions present per formula unit of the ionic salt is

\boxed{5}