An aqueous solution is prepared by dissolving 0.1 mol of an ionic salt in 1.8 kg of water at 35°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 °C is 60.00 mm of Hg. The number of ions present per formula unit of the ionic salt is_______ .

The correct answer is 5.

To find the number of ions present per formula unit of the ionic salt, we use Raoult's law for non-ideal solutions containing a solute that undergoes dissociation.

Step 1: Write down the given values
Moles of ionic salt, n_solute = 0.1 mol
Mass of water, w_water = 1.8 kg = 1800 g
Molar mass of water, M_water = 18 g/mol
Moles of water, n_water = 1800 / 18 = 100 mol
Degree of dissociation of the salt, α = 90% = 0.90
Vapor pressure of pure water at 35 °C, P° = 60.00 mm of Hg
Vapor pressure of the solution, P = 59.724 mm of Hg

Step 2: Relate relative lowering of vapor pressure to van 't Hoff factor (i)
According to Raoult's law for a dilute solution containing a dissociating solute:

$\frac{P^{°}-P}{P^{°}}=\frac{i·n_{\text{solute}}}{n_{\text{water}}+i·n_{\text{solute}}}$

Since n_water (100 mol) is much greater than i · n_solute, we can approximate the total moles in the denominator as n_water:

$\frac{P^{°}-P}{P^{°}}=\frac{i·n_{\text{solute}}}{n_{\text{water}}}$

Step 3: Calculate the van 't Hoff factor (i)
Substitute the given values into the formula:

$\frac{60.00-59.724}{60.00}=\frac{i·0.1}{100}$

$\frac{0.276}{60.00}=\frac{i·0.1}{100}$

$0.0046=0.001·i$

$i=\frac{0.0046}{0.001}=4.6$

Step 4: Determine the number of ions per formula unit (x)
The relation between van 't Hoff factor (i), degree of dissociation (α), and number of ions per formula unit (x) is:

$i=1+(x-1)\alpha$

Substitute i = 4.6 and α = 0.90:

$4.6=1+(x-1)·0.90$

$3.6=(x-1)·0.90$

$x-1=\frac{3.6}{0.90}=4$

$x=5$

Therefore, the number of ions present per formula unit of the ionic salt is 5.