Chapter 11, Problem 110QP

Interpretation Introduction

Interpretation:

The molarity of the sulfuric acid solution is to be determined.

Concept Introduction:

Molarity $\left(M\right)$ is defined as the number of moles $\left(n\right)$ of solute present in the volume of solution $\left(V\right)$ in litres. Molarity describes the concentration in terms of the volume of the solution.

$M=\frac{n}{V}$ …… (1)

Percent by mass concentration is the amount (in grams) of solute present in grams of solution multiplied by 100%.

$\%\text{ mass}=\frac{m_{\text{solute}}\left(\text{g}\right)}{m_{\text{solution}}\left(\text{g}\right)}\times 100$ …… (2)

Answer to Problem 110QP

Solution:

The molarity of given sulfuric acid solution is $2.865{\text{ mol L}}^{-1}$ .

Explanation of Solution

Given Information:

Percent by mass concentration of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ solution is $24\%$ and density of solution is $1.17{\text{ g mL}}^{-1}$ .

Substitute mass percentage as $24\%$ and mass of solution as $100\text{ g}$ in equation (2) as follows:

$\begin{array}{c}24\%=\frac{m_{\text{solute}}}{100\text{ g}}\times 100\\ m_{\text{solute}}=24\text{ g}\end{array}$

$100\text{ g}$ of solution contains $24\text{ g}$ sulfuric acid. The density of the solution is $\text{1}{\text{.17 g mL}}^{-1}$ .

The density $\left(\rho \right)$ of the solution is equal to the ratio of mass $\left(m\right)$ of solution to the volume $\left(V\right)$ of the solution.

$\rho =\frac{m}{V}$ …… (3)

Substitute $\rho$ as $\text{1}{\text{.17 g mL}}^{-1}$ and $V$ as $\text{1mL}$ in equation (3) as follows:

$\begin{array}{c}\text{1}{\text{.17 g mL}}^{-1}=\frac{m}{\text{1 mL}}\\ m=\text{1}{\text{.17 g mL}}^{-1}\times \text{1mL}\\ \text{=1}\text{.17 g}\end{array}$

$\text{1mL}$ solution weighs $\text{1}\text{.17}\text{g}$ .

$\begin{array}{c}1.17\text{ g}=1\text{ mL}\\ \text{1 g}=\frac{1\text{ mL}}{1.17\text{ g}}\end{array}$

Thus, the volume of $100\text{ g}$ solution is,

$\begin{array}{c}\text{100 g}=\frac{1\text{ mL}}{1.17\text{ g}}\times \text{100 g}\\ =85.47\text{ mL}\end{array}$

Now, $85.47\text{ mL}$ contains $24{\text{ g H}}_2{\text{SO}}_4$ .

$\begin{array}{c}85.47\text{ mL}=24{\text{ g H}}_2{\text{SO}}_4\\ 1\text{ mL}=\frac{24{\text{ g H}}_2{\text{SO}}_4}{85.47\text{ mL}}\end{array}$

So, $1000{\text{ mL H}}_2{\text{SO}}_4$ contains:

$\begin{array}{c}1000\text{ mL}=\frac{24{\text{ g H}}_2{\text{SO}}_4}{85.47\text{ mL}}\times 1000\text{ mL}\\ =\text{280}\text{.80 g}\end{array}$

Thus, the weight of Sulphuric acid in $1000{\text{ mL H}}_2{\text{SO}}_4$ of solution is $\text{280}\text{.80 g}$ . The number of moles $\left(n\right)$ of solute is equal to the ratio of mass $\left(m\right)$ of solute in solution to the molar mass $\left(MM\right)$ of solute.

$n=\frac{m}{MM}$ …… (4)

The molar mass of ${\text{H}}_2{\text{SO}}_4$ is as follows:

$\begin{array}{c}{\text{H}}_2{\text{SO}}_4=\left(\text{2}\times {\text{1 g mol}}^{-1}\right)+{\text{ 32 g mol}}^{-1}+\left(\text{4}\times {\text{16 g mol}}^{-1}\right)\\ =\text{ 2}{\text{g mol}}^{-1}+\text{32}{\text{g mol}}^{-1}+\text{64}{\text{g mol}}^{-1}\\ =\text{ 98}{\text{g mol}}^{-1}\end{array}$

Substitute $MM$ as $\text{98}{\text{g mol}}^{-1}$ and $m$ as $\text{280}\text{.80 g}$ in equation (4) as follows:

$\begin{array}{c}n=\frac{\text{280}\text{.80 g}}{\text{98}{\text{g mol}}^{-1}}\\ =2.865\text{ mol}\end{array}$

Now, substitute $n$ as $2.865\text{ mol}$ and $V$ as $1\text{ L}$ in equation (1) as follows:

$\begin{array}{c}M=\frac{2.865\text{mol}}{1\text{L}}\\ =2.865{\text{ mol L}}^{-1}\end{array}$ $$

Conclusion

The molarity of the sulfuric acid solution is $2.865{\text{ mol L}}^{-1}$ .