Calculate Number of Ions per Formula Unit of Ionic Salt

To determine the number of ions present per formula unit of the ionic salt, we use the principle of relative lowering of vapour pressure for a solution containing a non-volatile electrolyte.

1. Given Data:

• Moles of ionic salt (solute), $n_{\text{solute}} = 0.1 \text{ mol}$
• Mass of water (solvent), $w_{\text{water}} = 1.8 \text{ kg} = 1800 \text{ g}$
• Molar mass of water, $M_{\text{water}} = 18 \text{ g mol}^{-1}$
• Moles of water, $n_{\text{water}} = \frac{1800}{18} = 100 \text{ mol}$
• Degree of dissociation of the salt, $\alpha = 90\% = 0.90$
• Vapour pressure of pure water, $P^\circ = 60.000 \text{ mm of Hg}$
• Vapour pressure of the solution, $P = 59.724 \text{ mm of Hg}$

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2. Calculation of van 't Hoff Factor ($i$):

The relative lowering of vapour pressure is given by Raoult's Law: $\frac{P^\circ - P}{P^\circ} = \frac{i \cdot n_{\text{solute}}}{n_{\text{water}}}$

Substitute the given values into the equation: $\frac{60.000 - 59.724}{60.000} = \frac{i \times 0.1}{100}$

$\frac{0.276}{60.000} = \frac{0.1 \cdot i}{100}$

$0.0046 = \frac{0.1 \cdot i}{100}$

$0.1 \cdot i = 0.46$

$i = 4.6$

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3. Determination of Number of Ions per Formula Unit ($x$):

Let $x$ be the number of ions produced per formula unit of the ionic salt upon complete dissociation.

The relationship between the van 't Hoff factor ($i$), degree of dissociation ($\alpha$), and the number of ions ($x$) is: $i = 1 + (x - 1)\alpha$

Substitute $i = 4.6$ and $\alpha = 0.90$: $4.6 = 1 + (x - 1) \times 0.90$

$3.6 = 0.90 \cdot (x - 1)$

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

$x = 5$

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Final Answer:

The number of ions present per formula unit of the ionic salt is $5$.