What would be the molality of 20% (mass/mass) aqueous solution of KI? (molar mass of KI = 166 g mol$$^{-1}$$)

We begin by interpreting the phrase “20 % (mass/mass) aqueous solution of KI”. By definition, a 20 % (w/w) solution contains 20 g of solute (here, KI) in every 100 g of the entire solution.

So for 100 g of the solution:

$$\text{mass of KI (solute)} = 20\ \text{g}$$

$$\text{mass of water (solvent)} = 100\ \text{g} - 20\ \text{g} = 80\ \text{g}$$

Next, we need the number of moles of KI present. We recall the formula

$$\text{Moles of solute},\ n = \frac{\text{mass of solute}}{\text{molar mass}}.$$

Substituting the given values, we obtain

$$n = \frac{20\ \text{g}}{166\ \text{g mol}^{-1}}.$$

Carrying out the division,

$$n = 0.12048\ \text{mol}.$$

The molality $$m$$ of a solution is defined as

$$m = \frac{\text{moles of solute}}{\text{mass of solvent in kilograms}}.$$

First, we express the solvent mass in kilograms:

$$80\ \text{g} = 80 \times 10^{-3}\ \text{kg} = 0.08\ \text{kg}.$$

Now we substitute into the molality formula:

$$m = \frac{0.12048\ \text{mol}}{0.08\ \text{kg}}.$$

Dividing, we get

$$m = 1.506\ \text{mol kg}^{-1}.$$

On rounding to three significant figures,

$$m \approx 1.51\ \text{mol kg}^{-1}.$$

Hence, the correct answer is Option B.