Chapter 6, Problem 6.36P

Interpretation Introduction

(a)

Interpretation:

The preparation of 250 mL of 3.6% w/v solution of NaCl in water should be explained.

Concept Introduction:

The concentration is shown as w/v meaning this is mass of solute per volume of solution.

Answer to Problem 6.36P

$\text{9 g}$ of the solute must be added.

Explanation of Solution

Given Information:

$\begin{array}{l}V\text{}=250.\text{0 mL}\\ \text{C}\text{}=\text{}3.6\%\text{w/v}\end{array}$

Concentration has several ways to be calculated. It typically relates an amount of solute and the solution. Then given in mass of solute per unit volume of solution, or w/w, the total mass of the solute, solvent and solution must be known.

The formula for w/v is given as:

${\text{C}}_{w/v}=\frac{m_{\mathrm{solute}}}{V_{\text{solution}}}\times 100\%$.

Substitute known data and solve for mass of solute.

$\begin{array}{l}{\text{C}}_{w/v}=\frac{m_{\mathrm{solute}}}{V_{\text{solution}}}\times 100\%\\ 3.6\%w/v=\frac{m_{\mathrm{solute}}}{25\text{0 mL}}\times 100\%\\ m_{\mathrm{solute}}=\frac{3.6\%w/v}{100\%}\times 25\text{0 mL}\\ m_{\mathrm{solute}}=9\text{ g}\end{array}$

Thus, $9\text{ g}$ of solute is added to the remaining volume of solvent.

Interpretation Introduction

(b)

Interpretation:

The preparation of solution should be explained.

Concept Introduction:

The concentration is shown as w/v meaning this is mass of solute per volume of solution.

Answer to Problem 6.36P

$30.62\text{5 g}$ of the solute must be added.

Explanation of Solution

Given:

$\begin{array}{l}V\text{}=625.\text{0 mL}\\ \text{C}\text{}=\text{}4.9\%\text{w/v}\end{array}$

Concentration has several ways to be calculated. It typically relates an amount of solute and the solution. Then given in mass of solute per unit volume of solution, or w/w, the total mass of the solute, solvent and solution must be known.

The formula for w/v is given as:

${\text{C}}_{w/v}=\frac{m_{\mathrm{solute}}}{V_{\text{solution}}}\times 100\%$.

Substitute known data and solve for mass of solute.

$\begin{array}{l}{\text{C}}_{w/v}=\frac{m_{\mathrm{solute}}}{V_{\text{solution}}}\times 100\%\\ 4.9\%\text{w/v}=\frac{m_{\mathrm{solute}}}{625.\text{0 mL}}\times 100\%\\ m_{\mathrm{solute}}=\frac{3.6\%\text{w/v}}{100\%}\times 25\text{0 mL}\\ m_{\mathrm{solute}}=\text{30}\text{.625 g}\end{array}$

Thus, $\text{30}\text{.625 g}$ of solute is added to the remaining volume of solvent.

(c)

Interpretation Introduction

Interpretation:

To calculate how to create the solution with the given concentration and volume.

Concept Introduction:

The concentration is shown as w/v meaning this is mass of solute per volume of solution.

Answer to Problem 6.36P

$5.9595g$ of the solute must be added.

Explanation of Solution

Given:

$\begin{array}{l}V\text{}=43.5\text{m}\text{L}\\ \text{C}\text{}=\text{}13.7\%w/v\end{array}$

Concentration has several ways to be calculated. It typically relates an amount of solute and the solution. Then given in mass of solute per unit volume of solution, or w/w, the total mass of the solute, solvent and solution must be known.

The formula for w/w is given as:

${\text{C}}_{w/v}=\frac{m\text{a}ssofsolute}{volum\text{e}ofso\text{l}ution}\times 100\%$

Substitute known data and solve for mass of solute.

$\begin{array}{l}{\text{C}}_{w/v}=\frac{m\text{a}ssofsolute}{volum\text{e}ofso\text{l}ution}\times 100\%\\ 13.7\%v/v=\frac{m\text{a}ssofsolute}{43.5\text{m}\text{L}}\times 100\%\\ M_{solute}=\frac{13.7\%v/v}{100\%}\times (250\text{m}\text{L})\\ M_{solute}=5.9595g\end{array}$

Then, add $5.9595g$ to then add the remaining volume of solvent up to the required volume mark.

Conclusion

The solution should include solute and solvent amounts.

Interpretation Introduction

(d)

Interpretation:

To calculate how to create the solution with the given concentration and volume.

Concept Introduction:

The concentration is shown as w/v meaning this is mass of solute per volume of solution.

Answer to Problem 6.36P

$10.878g$ of the solute must be added.

Explanation of Solution

Given:

$\begin{array}{l}V\text{}=518\text{m}\text{L}\\ \text{C}\text{}=2.1\%w/v\end{array}$

Concentration has several ways to be calculated. It typically relates an amount of solute and the solution. Then given in mass of solute per unit mass of solvent, or w/w, the total mass of the solute, solvent and solution must be known.

The formula for w/w is given as:

${\text{C}}_{w/v}=\frac{m\text{a}ssofsolute}{volum\text{e}ofso\text{l}ution}\times 100\%$

Substitute known data and solve for mass of solute.

$\begin{array}{l}{\text{C}}_{w/v}=\frac{m\text{a}ssofsolute}{volum\text{e}ofso\text{l}ution}\times 100\%\\ 2.1\%v/v=\frac{m\text{a}ssofsolute}{518\text{m}\text{L}}\times 100\%\\ M_{solute}=\frac{2.1\%v/v}{100\%}\times (518\text{m}\text{L})\\ M_{solute}=10.878g\end{array}$

Then, add $10.878g$ to then add the remaining volume of solvent up to the required volume mark.

Conclusion

The solution should include solute and solvent amounts.