Chapter 21, Problem 3STP

Interpretation Introduction

Interpretation:

The molality of solution needs to be determined that contains $\text{0}\text{.25 g}$ of dichlorobenzene ( ${\text{C}}_{\text{6}}{\text{H}}_4{\text{Cl}}_2$ ) dissolved in $10\text{ g}$ of cyclohexane ( ${\text{C}}_{\text{6}}{\text{H}}_{12}$ ).

Concept introduction:

The concentration can be defined in terms of molality, which is a ratio of moles of solute present in mass of solvent (in kg). The mathematical formula is:

$\text{molality = }\frac{\text{number of moles of solute}}{\text{Mass of solvent (in kg)}}$ - (1)

Answer to Problem 3STP

Option (a), $\text{0}\text{.17 mol/kg}$.

Explanation of Solution

In solution, the substance which is present in lesser amount is said to be solute whereas the substance present in larger amount is said to be solvent. So, in the given case, ${\text{C}}_{\text{6}}{\text{H}}_4{\text{Cl}}_2$ is solute and ${\text{C}}_{\text{6}}{\text{H}}_{12}$ is solvent.

Now, calculating the number of moles of solute, ${\text{C}}_{\text{6}}{\text{H}}_4{\text{Cl}}_2$ using formula:

$\text{number of moles = }\frac{\text{given mass}}{\text{Molar mass}}$

Molar mass of ${\text{C}}_{\text{6}}{\text{H}}_4{\text{Cl}}_2$ is $\text{147}\text{.01 g/mol}$.

Substituting the values:

$\begin{array}{l}{\text{n}}_{{\text{C}}_{\text{6}}{\text{H}}_4{\text{Cl}}_2}\text{ = }\frac{\text{0}\text{.25 g}}{\text{147}\text{.01 g/mol}}\\ {\text{n}}_{{\text{C}}_{\text{6}}{\text{H}}_4{\text{Cl}}_2}\text{ = 0}\text{.0017 mol}\end{array}$

Converting mass of solvent, ${\text{C}}_{\text{6}}{\text{H}}_{12}$ from grams to kilograms as:

${\text{m}}_{{\text{C}}_{\text{6}}{\text{H}}_{12}\text{(in kg})}=10\text{ g}\times \frac{\text{1 kg}}{1000\text{ g }}\text{ = 0}\text{.01 kg}$

Now, the molality is calculated using equation (1) as:

$\text{molality = }\frac{{\text{n}}_{{\text{C}}_{\text{6}}{\text{H}}_4{\text{Cl}}_2}}{{\text{m}}_{{\text{C}}_{\text{6}}{\text{H}}_{12}\text{(in kg})}}$

Substituting the values as:

$\begin{array}{l}\text{molality = }\frac{\text{0}\text{.0017 mol}}{\text{0}\text{.01 kg}}\\ \text{molality = 0}\text{.17 mol/kg = 0}\text{.17 m}\end{array}$

Hence, the molality of solution that contains $\text{0}\text{.25 g}$ of dichlorobenzene ( ${\text{C}}_{\text{6}}{\text{H}}_4{\text{Cl}}_2$ ) dissolved in $10\text{ g}$ of cyclohexane ( ${\text{C}}_{\text{6}}{\text{H}}_{12}$ ) is $\text{0}\text{.17 mol/kg}$.