Question

An aqueous solution is $\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{\%}\mathbf{\hspace{0.17em}}\mathbf{NaCl}$by mass and has a density of $\mathbf{1}\mathbf{.}\mathbf{071}\mathbf{}\raisebox{1ex}{$\mathbf{g}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{c}{\mathbf{m}}^{\mathbf{3}}$}\right.$at $\mathbf{25}\mathbf{°}\mathbf{C}$. The observed osmotic pressure of this solution is $\mathbf{7}\mathbf{.}\mathbf{83}\mathbf{}\mathbf{atm}$ at $\mathbf{25}\mathbf{°}\mathbf{C}$.

a. What fraction of the moles of $\mathbf{NaCl}$in this solution exist as ion pairs?

b. Calculate the freezing point that would be observed for this solution.

Short answer

$i=0.875$

Freezing point$=0.5927°\mathrm{C}$

Step-by-step solution

calculation of the volume of solution

Assume $100\mathrm{g}$ of solution.

So, mass of $\mathrm{NaCl}=1\mathrm{g}$ because the solution is $1.00\%\mathrm{NaCl}$ by mass.

Calculate volume of solution using density: we know that density$=\frac{mass}{volume}$

Volume of solution=$\frac{\mathrm{mass}}{\mathrm{density}}=\frac{100}{1.071}=93.37\mathrm{ml}$

Calculate moles and molarity of sodium chloridemolar mass=58.44

Moles of $\mathrm{NaCl}=\frac{\mathrm{given}\mathrm{mass}}{\mathrm{molar}\mathrm{mass}}=\frac{1\mathrm{g}}{58.44{\mathrm{gmol}}^{-1}}=0.017\mathrm{moles}$

$\mathrm{Molarity}=\frac{\mathrm{moles}}{\mathrm{volume}\mathrm{of}\mathrm{solution}\mathrm{in}\mathrm{L}}=\frac{0.017\mathrm{mol}}{0.09337\mathrm{L}}=0.182\mathrm{M}$

Calculate i

We know that osmotic pressure$=\mathrm{MRTi}$

Thus,

$$\begin{gathered} 7.83=0.182\times 0.08205\times 298\times \mathrm{i} \\ \mathrm{i}=1.751 \end{gathered}$$

Since we have $2$ particles in solution, sodium and chloride ion, therefore, the fraction of ion pairs

$\mathrm{i}=\frac{1.751}{2}=0.875$

Calculate the freezing point

We know that,${\mathrm{K}}_{\mathrm{f}}$of water is equal to$1.86\raisebox{1ex}{$°\mathrm{C}·\mathrm{kg}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mole}$}\right.$.

Now, the depression in the freezing point

$∆{\mathrm{Tf}}_{\mathrm{f}}=\mathrm{i}\times {\mathrm{K}}_{\mathrm{f}}\times \mathrm{M}=1.751\times 1.86\times 0.182\mathrm{M}=0.5927°\mathrm{C}$

Now, freezing point of water$=0°C$
the freezing point of the solution =$0+0.5927°\mathrm{C}$