Question

The expressions for boiling-point elevation and freezing point depression apply accurately to dilute solutions only. A saturated aqueous solution of $\mathbf{\text{NaI}}$(sodium iodide) in water has a boiling point of$\mathbf{\text{144°C}}$.The mole fraction of$\mathbf{\text{NaI}}$in the solution is$\mathbf{\text{0.390}}$. Compute the molality of this solution. Compare the boiling-point elevation predicted by the expression in this chapter with the elevation actually observed.

Short answer

The molality of the solution is 35.5 m.

The boiling-point elevation obtained is ${36.5}^{°}\text{C}$. The observed elevation is ${44}^{°}\text{C}$, which may be because of the concentrated solution.

Step-by-step solution

Given information

The equation known for the boiling point elevation is ${\mathbf{\Delta T}}_b\mathbf{=}{\mathbf{K}}_b\mathbf{m}$.

Where,

• ${\mathbf{K}}_b$= Molal boiling point constant
• m = Molality of solution
• ${\mathbf{\Delta T}}_b$= Change in boiling point

$K_{\text{b}}$for water is${0.512}^{°}{\text{Ckgmol}}^{-1}$

The boiling point of sodium iodide is ${144}^{°}\text{C}$.

${144}^{°}\text{C}-{100}^{°}\text{C}={44}^{°}\text{C}$

Calculation of the molar solute concentration

Substitute the values in the equation:

$\begin{array}{c}{44}^{°}\text{C}={0.512}^{°}{\text{Ckgmol}}^{-1}\mathrm{m}\\ \mathrm{m}={\frac{{44}^{°}\text{C}}{{0.512}^{°}\text{Ckgmol}}}^{-1}\\ =85.9\text{mol}/\text{kg}\\ \frac{85.9\text{mol}/\text{kg}}{2}=42.95\text{mol}/\text{kg}\end{array}$

Which is around $\approx 43.0\text{mol}/\text{kg}$.

Calculate the mass of water

The molar mass of water is 18.02 g/ mol.

Mass of water $=0.610\text{mol}\times 18.02\text{g}{\text{mol}}^{-1}=10.99\text{g}\approx 11.0\text{g}\approx 0.0110\text{kg}$.

Calculation of molality

Molality is measured by the number of sodium iodide fraction in the water:

Change in boiling point elevation

The boiling point change is calculated by formula:$\Delta T_{\text{b}}=K_{\text{b}}m$

$\begin{array}{c}{\mathrm{\Delta T}}_{\text{b}}=\left({0.512}^{°}{\text{Ckgmol}}^{-1}\right)2\left(35.5{\text{molkg}}^{-1}\right)\\ ={36.5}^{°}\text{C}\end{array}$