Chapter 13, Problem 123E

Interpretation Introduction

Interpretation:

The mass of glucose which needs to be added in the $150\text{ mL}$ water is to be calculated.

Concept introduction:

The relation between numbers of moles $\left(n\right)$ and molar mass is:

$n=\frac{m}{\text{molar mass}}$

Here,$m$ is mass

The number of moles of solute to the mass of solvent is called molality.

The mass of the solvent should be in kilograms $\left(\text{kg}\right)$.

$m=\frac{\text{Moles solute}}{\text{Kilograms solvent}}$

Freezing point is the temperature at which change of liquid state to solid state.

The freezing point depression is as follows:

$\Delta T_{\text{f}}=m\times K_{\text{f}}$

Here, $\Delta T_{\text{f}}$ is the change in temperature of the freezing point in $\text{C}_{\circ }$,$m$ is the

molality of the solution in $\frac{\text{mol solute}}{\text{kg solvent}}$ and $K_{\text{f}}$ is the freezing point depression

constant for the solvent.

For water:

$K_{\text{f}}=1.86\frac{\text{C}\circ \text{ kg solvent}}{\text{mol solute}}$