Chapter 16, Problem 78A

Interpretation Introduction

Interpretation : The mole fraction of ${\text{NaHCO}}_{\text{3}}$ in a saturated solution and the molality of the solution are to be calculated.

Concept Introduction : The molality and mole fraction are the two ways to express the percentage of solute to solvent particles.

The formula of the mole fraction is given as:

  $\begin{array}{l}{X}_{\text{A}}=\frac{n_{\text{A}}}{n_{\text{A}}+n_{\text{B}}}\\ X_{\text{B}}=\frac{n_{\text{B}}}{n_{\text{A}}+n_{\text{B}}}\end{array}$

Here, $X_{\text{A}}$ is the mole fraction of solute A, $X_{\text{B}}$ is the mole fraction of solute B, $n_{\text{A}}$ is the mole of solute A, and $n_{\text{B}}$ is the mole of solute B.

The formula of molality is given as:

  $\text{molality}\left(\text{m}\right)=\frac{\text{moles}\text{of}\text{solute}}{\text{kilogram}\text{of}\text{solvent}}$

Answer to Problem 78A

The mole fraction of ${\text{NaHCO}}_{\text{3}}$

  $0.02$ .

The molality of ${\text{NaHCO}}_{\text{3}}$ is $1.14\text{mole/kg}$ .

Explanation of Solution

The given mass of ${\text{NaHCO}}_{\text{3}}$ is $9.6\text{g}$ .

The given mass of water is $100\text{g}$ .

The molar mass of ${\text{NaHCO}}_{\text{3}}$ is $84\text{g/mol}$ .

The molar mass of water is $18\text{g/mol}$ .

The formula to calculate the moles is given as:

  $\text{moles}\text{=}\frac{\text{given}\text{mass}}{\text{molar}\text{mass}}$   ...... (1)

To calculate the moles of ${\text{NaHCO}}_{\text{3}}$ , substitute the values in equation (1).

  $\begin{array}{c}\text{moles}\text{=}\frac{\text{given}\text{mass}}{\text{molar}\text{mass}}\\ =\frac{9.6\text{g}}{84\text{g/mol}}\\ =0.114\text{moles}\end{array}$

To calculate the moles of water, substitute the values in equation (1).

  $\begin{array}{c}\text{moles}\text{=}\frac{\text{given}\text{mass}}{\text{molar}\text{mass}}\\ =\frac{100\text{g}}{18\text{g/mol}}\\ =5.55\text{moles}\end{array}$

The formula of the mole fraction of ${\text{NaHCO}}_{\text{3}}$ is given as:

  $X_{{\text{NaHCO}}_{\text{3}}}=\frac{n_{{\text{NaHCO}}_{\text{3}}}}{n_{{\text{NaHCO}}_{\text{3}}}+n_{{\text{H}}_2\text{O}}}$   ...... (2)

To calculate the mole fraction of ${\text{NaHCO}}_{\text{3}}$, substitute the values in formula (2) as follows:

  $\begin{array}{c}{X}_{{\text{NaHCO}}_{\text{3}}}=\frac{n_{{\text{NaHCO}}_{\text{3}}}}{n_{{\text{NaHCO}}_{\text{3}}}+n_{{\text{H}}_2\text{O}}}\\ =\frac{0.114}{0.114+5.55}\\ =0.02\end{array}$

The mole fraction of ${\text{NaHCO}}_{\text{3}}$

  $0.02$ .

The formula of molality is given as

  $\text{molality}\left(\text{m}\right)=\frac{\text{moles}\text{of}\text{solute}}{\text{kilogram}\text{of}\text{solvent}}$

To calculate the molality of ${\text{NaHCO}}_{\text{3}}$, substitute the values in the above formula:

  $\begin{array}{c}\text{molality}\left(\text{m}\right)=\frac{\text{moles}\text{of}\text{solute}}{\text{kilogram}\text{of}\text{solvent}}\\ =\frac{0.114\text{mole}}{100\text{g}}\left(1\text{g}\text{=}{\text{10}}^{-3}\text{kg}\right)\\ =\frac{0.114\text{mole}}{0.1\text{kg}}\\ \text{m}=1.14\text{mole/kg}\end{array}$

The molality of ${\text{NaHCO}}_{\text{3}}$ is $1.14\text{mole/kg}$ .