Chapter 16, Problem 70A

Interpretation Introduction

Interpretation : The aqueous solution of methanol in which the mole fraction of methanol is $0.40$ is to be described.

Concept Introduction : The molality and mole fraction are the two ways to express the percentage of solute to solvent particles. The ratio of a solute's moles to the sum of the moles of solvent and solute in a solution is known as the solute's mole fraction.

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.

Answer to Problem 70A

The aqueous solution of methanol is prepared by mixing $\text{32}\text{g}$ of methanol with $27\text{g}$ of water.

Explanation of Solution

The mole fraction of methanol is $0.40$ .

The molar mass of methanol is $32\text{g/mol}$ .

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

The moles of methanol $\left(n_{{\text{CH}}_3\text{OH}}\right)$ is $1$

The formula of the mole fraction of methanol with water as a solvent is given as:

  $X_{{\text{CH}}_{\text{3}}\text{OH}}=\frac{n_{{\text{CH}}_{\text{3}}\text{OH}}}{n_{{\text{CH}}_{\text{3}}\text{OH}}+n_{{\text{H}}_2\text{O}}}$   ...... (1)

The moles of water are to be found.

To calculate the moles of water, substitute the values in formula (1) as follows:

  $\begin{array}{c}{X}_{{\text{CH}}_{\text{3}}\text{OH}}=\frac{n_{{\text{CH}}_{\text{3}}\text{OH}}}{n_{{\text{CH}}_{\text{3}}\text{OH}}+n_{{\text{H}}_2\text{O}}}\\ 0.40=\frac{1}{1+n_{{\text{H}}_2\text{O}}}\\ 0.40+0.40n_{{\text{H}}_2\text{O}}=1\\ 0.40n_{{\text{H}}_2\text{O}}=1-0.40\end{array}$

Further, solve the above equation.

  $\begin{array}{c}0.40n_{{\text{H}}_2\text{O}}=1-0.40\\ 0.40n_{{\text{H}}_2\text{O}}=0.60\\ n_{{\text{H}}_2\text{O}}=\frac{0.60}{0.40}\\ =1.5\end{array}$

The moles of water are $1.5$ .

To calculate the amounts of solute and solvent, multiply the moles by their molar mass.

The amount of ${\text{CH}}_3\text{OH}$ is given as

  $\begin{array}{c}\text{Mass}\text{=}1\text{mol}\times \text{32}\text{g/mol}\\ \text{=}\text{32}\text{g}\end{array}$

The amount of ${\text{H}}_2\text{O}$ is given as

  $\begin{array}{c}\text{Mass}\text{=}1.5\text{mol}\times 18\text{g/mol}\\ \text{=}27\text{g}\end{array}$

The aqueous solution of methanol is prepared by mixing $\text{32}\text{g}$ of methanol with $27\text{g}$ of water.