Question

An aqueous solution containing 20.0wt%KI has a density of1.168g/mL. Find the molality (m, not M) of the KI solution.

Short answer

The molality of the KI solution is $1.51\mathbf{mol}\mathbf{/}\mathbf{kg}$

Step-by-step solution

Definition of molality.

The number of moles of solute in a solution equivalent to $1kg$or $1000g$of asolvent is known as its molality. The definition of molarity, on the other hand, is based on a certain volume of solution.

The molality of KI solution.

To begin, remember that molality, expressed as $m1$equals:

$m1=\frac{\mathbf{moles}\mathbf{}\mathbf{of}\mathbf{}\mathbf{solute}\left(\mathrm{KI}\right)}{\mathbf{mass}\mathbf{}\mathbf{of}\mathbf{}\mathbf{solvent}\left(\mathrm{water}\right)}$

It is given that the density of $1.168\mathbf{g}\mathbf{/}\mathbf{mL}$which is $1.168\mathbf{g}$of KI in $1\mathbf{mL}$of water, is $1168\mathbf{g}$in 1L

The total mass of the solute (KI) is equal to$0.2\times 1168\mathbf{g}=233.6\mathbf{g}$

The total mass of the solvent (water) is equal to $0.8\times 1168\mathbf{g}=934.4\mathbf{g}$

Now, calculate the number of moles for KI:

$$\begin{gathered} n\left(\mathbf{KI}\right)=\frac{\mathrm{m}}{\mathrm{M}} \\ =\frac{233.6\mathbf{g}}{166\mathbf{g}\mathbf{/}\mathbf{mol}} \\ =1.407\mathbf{mol} \end{gathered}$$

Now, calculate the molality $\left(m1\right)$

$$\begin{gathered} m1=\frac{\mathrm{n}\left(\mathbf{KI}\right)}{\mathrm{m}\left(\mathbf{water}\right)} \\ =\frac{1.407\mathbf{mol}}{934.4\mathbf{g}} \\ =1.51\times {10}^{-3}\mathbf{mol}\mathbf{/}\mathbf{g} \\ =1.51\mathbf{mol}\mathbf{/}\mathbf{kg} \end{gathered}$$

Hence, the molality is $1.51\mathbf{mol}\mathbf{/}\mathbf{kg}$