Chapter 16, Problem 92A

Interpretation Introduction

Interpretation: The mole fractions of sugar and water are to be calculated in which the molality of that corresponding solution is given as $1.62\text{m}$ .

Concept introduction: Mole fraction is a term that can be used to express solutions' concentration. The mole fraction of a particular solute is the ratio of the number of moles of solute present in the solution to the total number of moles present in the solution is the addition of the number of moles of solute and solvent.

The formula that can be used for the calculation of mole fraction is,

  $X_a=\frac{n_a}{n_a+n_b}$

Here, $n_a$ are the moles of solvent, $n_b$ are the moles of solute, and $X_a$ is the mole fraction of solute.

And,

  $X_b=\frac{n_b}{n_a+n_b}$

Here, $n_a$ are the moles of solvent, $n_b$ are the moles of solute, and $X_b$ is the mole fraction of solvent.

The expression for moles is as follows:

  $\text{moles}=\frac{\text{ mass of solute}}{\text{Molar mass of solute}}$   ...... (1)

Answer to Problem 92A

Mole fraction of water is $0.97$ .

Mole fraction of glucose is $0.021$

Explanation of Solution

The molality of a solution is the ratio of moles of solute to the mass of solvent in kilogram.

The molality of the glucose solution is given as $1.62\text{m}$ . And this means that a $1.62\text{m}$ solution of glucose contains $1.62\text{moles}$ of glucose in $1\text{kg}\left(1000\text{g}\right)$ of water.

The molecular mass of water is 18.

Substitute, $1000\text{g}$ for the mass of water and $\text{ 18 g/mol}$ for molar mass of water in equation (1).

  $\begin{array}{c}\text{moles}\left({\text{H}}_2\text{O}\right)=\frac{\text{ 1000 g}}{\text{18 g/mol}}\\ =55.5\text{ mol}\end{array}$

Substitute $1.2$ for $n_a$ and $55.5$ for $n_b$ in the formula of mole fraction of solute to calculate the mole fraction of solute glucose, $X_a$ .

  $\begin{array}{c}{X}_a=\frac{1.2}{1.2+55.5\text{mol}}\\ =0.021\end{array}$

Substitute $1.2$ for $n_a$ and $55.5$ for $n_b$ in the formula of mole fraction of solvent to calculate the mole fraction of solvent water, $X_b$ .

  $\begin{array}{c}{X}_b=\frac{55.5}{1.2+55.5\text{mol}}\\ =0.97\end{array}$

Conclusion

Therefore,

Mole fraction of water is $0.97$ .

Mole fraction of glucose is $0.021$