Chapter 13, Problem 19QP

Interpretation Introduction

Interpretation:

The molality of each of the given aqueous solutions is to be calculated.

Concept introduction:

Molality is the ratio of thenumber of moles of solute to the mass of the solvent in $\text{kg}$.

The formula to calculate molality $\left(m\right)$ is as follows:

$m=\frac{\text{moles of solute}}{\text{mass of solvent}\left(\text{in kg}\right)}$`…… (1)

Moles $\left(n\right)$ are expressed in terms of mass and molar mass as follows:

$n=\frac{\text{mass}}{\text{molar mass}}$ …… (2)

Answer to Problem 19QP

Solution:

a) $\text{1}\text{.7 m}$

b) $\text{0}\text{.87 m}$

c) $\text{7}\text{.0 m}$

Explanation of Solution

a) $\text{1}\text{.22 M}$ sugar $\left(C_{12}{\text{H}}_{22}{\text{O}}_{11}\right)$ solution

The molarity of the sugar solution is $\text{1}\text{.22 M}$ thatmeans $\text{1}\text{.22 mol}$ of sugar is present in $1\text{ L}$ of solution.

The density of the solution is $1120\text{ g/L}$.

Calculate the mass of the solution as follows:

$\left(1\cancel{\text{L}}\times \frac{1120\text{ g}}{\cancel{\text{L}}}\right)=1120\text{ g}$

Rearrange the equation (2) for mass as follows:

$\text{mass}=n\times \text{molar mass}$

The molar mass of sugar $\left(C_{12}{\text{H}}_{22}{\text{O}}_{11}\right)$ is $\text{342 g/mol}$.

Substitute $\text{1}\text{.22 mol}$ for $n$ and $\text{342 g/mol}$ for molar mass in the above equation as follows:

$\begin{array}{c}\text{mass}=1.22\text{ mol}\times 342\text{ g/mol}\\ =\text{417}\text{.2 g}\end{array}$

Now, calculate the mass of the solvent as follows:

$\text{Mass of solvent}=\text{Mass of solution}-\text{mass of solute}$

Substitute $1120\text{ g}$ for the mass of solution and $\text{417}\text{.2 g}$ for the mass of solute in the above expression

$\begin{array}{c}\text{Mass of solvent}=1120\text{ g}-417.2\text{ g}\\ =702.8\text{ g}\end{array}$

The relation between $\text{g}$ and $\text{kg}$ is as follows:

$1\text{ g }={\text{10}}^{-3}\text{ kg}$

Convert $\text{702}\text{.8 g}$ to $\text{kg}$:

$\text{702}\text{.8 g}\times \frac{{10}^{-3}\text{ kg}}{1\text{ g}}=0.7028\text{ kg}$ $$

Substitute $\text{1}\text{.22 mol}$ for moles of solute and $0.7028\text{ kg}$ for the mass of the solvent in equation (1)

$\begin{array}{c}m=\frac{\text{1}\text{.22 mol}}{\text{0}\text{.7028 kg }}\\ =1.7\text{ m}\end{array}$

b) $\text{0}\text{.87 M}$ $\text{NaOH}$ solution.

The molarity of $\text{NaOH}$ solution is $\text{0}\text{.87 M}$, which means that $\text{0}\text{.87 mol}$ of $\text{NaOH}$ is present in $1\text{ L}$ of solution.

The density of the solution is $1040\text{ g/L}$.

The mass of the solution as follows:

$\left(1\cancel{\text{L}}\times \frac{1040\text{ g}}{\cancel{\text{L}}}\right)=1040\text{ g}$

Rearrange the equation (2) for mass as follows:

$\text{mass}=n\times \text{molar mass}$

The molar mass of $\text{NaOH}$ is $\text{40 g/mol}$.

Substitute $\text{0}\text{.87 M}$ for $n$ and $\text{40 g/mol}$ for molar mass in the above equation

$\begin{array}{c}\text{mass}=0.87\text{ mol}\times 40\text{ g/mol}\\ =34.8\text{ g}\end{array}$

Now, calculate the mass of the solvent as follows:

$\text{Mass of solvent}=\text{Mass of solution}-\text{mass of solute}$

Substitute $1040\text{ g}$ for the mass of the solution and $\text{34}\text{.8 g}$ for the mass of thesolute in the above expression

$\begin{array}{c}\text{Mass of solvent}=1040\text{ g}-34.8\text{ g}\\ =1005.2\text{ g}\end{array}$

The relation between $\text{g}$ and $\text{kg}$ is as follows:

$1\text{ g }={\text{10}}^{-3}\text{ kg}$

Convert $\text{702}\text{.8 g}$ to $\text{kg}$ as follows:

$\text{1005}\text{.2 g}\times \frac{{10}^{-3}\text{ kg}}{1\text{ g}}=1.0052\text{ kg}$ $$

Substitute $\text{0}\text{.87 mol}$ for moles of solute and $\text{1}\text{.0052 kg}$ for the mass of solvent in equation (1)

$\begin{array}{c}m=\left(\frac{\text{0}\text{.87 mol}}{\text{1}\text{.0052 kg }}\right)\\ =0.87\text{ m}\end{array}$

c) $\text{5}\text{.24 M}$ ${\text{NaHCO}}_{\text{3}}$ solution.

The molarity of ${\text{NaHCO}}_{\text{3}}$ solution is $\text{5}\text{.24 M}$, that is, $\text{5}\text{.24 mol}$ of ${\text{NaHCO}}_{\text{3}}$ is present in $1\text{ L}$ of solution.

The density of the solution is $1190\text{ g/L}$.

Calculate the mass of the solution as follows:

$\left(1\cancel{\text{L}}\times \frac{1190\text{ g}}{\cancel{\text{L}}}\right)=1190\text{ g}$

Rearrange the equation (2) for mass.

$\text{mass}=n\times \text{molar mass}$

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

Substitute $\text{5}\text{.24 mol}$ for $n$ and $\text{84 g/mol}$ for molar mass in the above equation

$\begin{array}{c}\text{mass}=5.24\text{ mol}\times 84\text{ g/mol}\\ =440.2\text{ g}\end{array}$

Now, calculate the mass of solvent using the expression shown below:

$\text{Mass of solvent}=\text{Mass of solution}-\text{mass of solute}$

Substitute $1190\text{ g}$ for the mass of the solution and $\text{440}\text{.2 g}$ for the mass of solute in the above expression

$\begin{array}{c}\text{Mass of solvent}=1190\text{ g}-440.2\text{ g}\\ =749.8\text{ g}\end{array}$

The relation between $\text{g}$ and $\text{kg}$ is as follows:

$1\text{ g }={\text{10}}^{-3}\text{ kg}$

Convert $\text{749}\text{.8 g}$ to $\text{kg}$ as follows:

$\left(\text{749}\text{.8 }\cancel{\text{g}}\times \frac{{10}^{-3}\text{ kg}}{1\cancel{\text{g}}}\right)=0.7498\text{ kg}$ $$

Substitute $\text{5}\text{.24 mol}$ for the moles of solute and $\text{0}\text{.7498 kg}$ for the mass of solvent in equation (1)

$\begin{array}{c}m=\left(\frac{\text{5}\text{.24 mol}}{\text{0}\text{.7498 kg }}\right)\\ =7.0\text{ m}\end{array}$