Calculate the molality of KI if the density of 20% (mass/mass) aqueous solution of Kl is 1.202 g mL-1.

(Molar mass of Kl is 166 g mol-1)

The correct option is 1.5 mol kg^-1.

Step-by-Step Explanation:

1. Understand the given data:
- Mass percent of potassium iodide (KI) in the aqueous solution = 20% (mass/mass)
- Molar mass of KI ($M_{\text{solute}}$) = 166 g mol^-1
- Density of the solution = 1.202 g mL^-1

2. Interpret 20% (mass/mass) aqueous KI solution:
A 20% by mass aqueous solution of KI means that 20 g of KI (solute) is present in 100 g of the solution.
- Mass of solute (KI), $w_{\text{solute}}=20\text{g}$
- Mass of solution = 100 g
- Therefore, the mass of solvent (water), $w_{\text{solvent}}=\text{Mass of solution}-\text{Mass of solute}=100\text{g}-20\text{g}=80\text{g}$

To use this in the molality formula, we convert the mass of the solvent to kilograms:
$w_{\text{solvent}}=\frac{80}{1000}\text{kg}=0.08\text{kg}$

3. Calculate the number of moles of solute (KI):
The number of moles ($n$) of KI is given by:
$n=\frac{w_{\text{solute}}}{M_{\text{solute}}}$

Substituting the values:
$n=\frac{20\text{g}}{166\text{g}{\text{mol}}^{-1}}\approx 0.1205\text{mol}$

4. Calculate the molality (m) of the solution:
Molality is defined as the number of moles of solute dissolved per kilogram of the solvent:
$m=\frac{\text{Moles of solute (KI)}}{\text{Mass of solvent (water) in kg}}$

Substituting the calculated values into the formula:
$m=\frac{0.1205\text{mol}}{0.08\text{kg}}$

Evaluating this division:
$m\approx 1.506\text{mol kg}{-1}^{}$

Rounding to a standard precision gives approximately 1.5 mol kg^-1.