Chapter 3, Problem 3.66PAE

3.64 How many grams of solute are present in each of these

solutions?

(a) 37.2 mL ofO.471 M HBr

(b) 113.0 L of 1.43 M Na₂CO₃

(c) 212 mL of 6.8 M CH₃COOH

(d) 1.3 × 10-4 L of 1.03 M H2S03

Interpretation Introduction

To determine:

The mass in grams of solute for 37.2 mL of 0.471 M $HBr$ solution.

Explanation of Solution

Molar mass of $HBr$ is:

$\begin{array}{c}M{W}_{HBr}=M{W}_H+M{W}_{Br}\\ =1.0+79.9\\ =80.9g/mol\end{array}$

Molarity is defined as the number of moles of solute ( $HBr$ ) in 1 L of solution.

The solution is 0.471 M, i.e. it contains 0.471 moles of $HBr$

per 1000 mL solution. Thus, for 37.2 mL solution, the number of moles would be:

$n=0.471\times \frac{37.2mL}{1000mL}=0.0175\text{}moles$

Now, number of moles is given by:

$\begin{array}{l}n=\frac{Massofthecompound}{Molarmass}\\ 0.0175moles=\frac{Massofthecompound}{80.9g/mol}\\ Massofthecompound=0.0175\times 80.9=1.42g\end{array}$

Interpretation Introduction

To determine:

The mass in grams of solute for 113.0 L of 1.43 M $N{a}_{2}C{O}_3$ solution.

Explanation of Solution

Molar mass of $N{a}_{2}C{O}_3$ is:

$\begin{array}{c}M{W}_{N{a}_{2}C{O}_3}=2M{W}_{Na}+M{W}_C+3M{W}_O\\ =2\times 23.0+12.0+3\times 16.0\\ =106g/mol\end{array}$

Molarity is defined as the number of moles of solute ( $N{a}_{2}C{O}_3$ ) in 1 L of solution.

The solution is 1.43 M, i.e. it contains 1.43 moles of $N{a}_{2}C{O}_3$

per 1 L solution. Thus, for 113.0 L solution, the number of moles would be:

$n=1.43\times \frac{113.0L}{1L}=161.59moles$

Now, number of moles is given by:

$\begin{array}{l}n=\frac{Massofthecompound}{Molarmass}\\ 161.59moles=\frac{Massofthecompound}{106g/mol}\\ Massofthecompound=161.59\times 106=17128.5g\end{array}$

Interpretation Introduction

To determine:

The mass in grams of solute for 21.2mL of 6.8 M $C{H}_{3}COOH$ solution.

Explanation of Solution

Molar mass of $C{H}_{3}COOH$ is:

$\begin{array}{c}M{W}_{N{a}_{2}C{O}_3}=2M{W}_C+4M{W}_H+2M{W}_O\\ =2\times 12.0+4\times 1.0+2\times 16.0\\ =60g/mol\end{array}$

Molarity is defined as the number of moles of solute ( $C{H}_{3}COOH$ ) in 1 L of solution.

The solution is 6.8 M, i.e. it contains 6.8 moles of $C{H}_{3}COOH$

per 1000mL solution. Thus, for 21.2mL solution, the number of moles would be:

$n=6.8\times \frac{21.2mL}{1000ml}=0.144moles$

Now, number of moles is given by:

$\begin{array}{l}n=\frac{Massofthecompound}{Molarmass}\\ 0.144moles=\frac{Massofthecompound}{60g/mol}\\ Massofthecompound=0.144\times 60=8.65g\end{array}$

Interpretation Introduction

To determine:

The mass in grams of solute for $1.3\times {10}^{-4}L$ of 1.03 M $H_{2}S{O}_3$ solution.

Explanation of Solution

Molar mass of $H_{2}S{O}_3$ is:

$\begin{array}{c}M{W}_{H_{2}S{O}_3}=2M{W}_H+M{W}_S+3M{W}_O\\ =2\times 1.0+32.1+3\times 16.0\\ =82.1g/mol\end{array}$

Molarity is defined as the number of moles of solute ( $H_{2}S{O}_3$ ) in 1 L of solution.

The solution is 1.03 M, i.e. it contains 1.03 moles of $H_{2}S{O}_3$ per 1L solution. Thus, for $1.3\times {10}^{-4}L$

solution, the number of moles would be:

$n=1.03\times \frac{1.3\times {10}^{-4}L}{1L}=1.34\times {10}^{-4}moles$

Now, number of moles is given by:

$\begin{array}{l}n=\frac{Massofthecompound}{Molarmass}\\ 1.34\times {10}^{-4}moles=\frac{Massofthecompound}{82.1g/mol}\\ Massofthecompound=1.34\times {10}^{-4}\times 82.1=0.011g\end{array}$

Conclusion

Therefore, knowing the molarity of a given solution, the number of moles can be calculated. Using the molar mass, the mass of the solute present in the solution can be calculated.