Chapter 24, Problem 1ASA

Interpretation Introduction

(a)

Interpretation:

The moles of NaOH solution is to be calculated.

Concept introduction:

Number of moles can be calculated as to multiply the molarity by the volume in liters.

It is expressed as,

$\text{Moles}=\text{Molarity}\text{×}\text{Volume}\text{of}\text{solution}\text{in}\text{liter}$

Answer to Problem 1ASA

The moles of NaOH solution is 0.0432 mol.

Explanation of Solution

Volume of initial NaOH solution, (V₁) = 7.2 mL= 0.0072 L

Molarity of NaOH solution, (M₁) = 6.0 M

Volume of final NaOH solution, (V₂) = 400.0 mL= 0.4 L

First we have to find the moles of NaOH, it is calculated as

$\text{Moles}=\text{Molarity}\text{×}\text{Volume}\text{of}\text{solution}\text{in}\text{liter}$

Substitute the values of molarity and volume of NaOH in the above expression.

$\begin{array}{rll}\text{Moles}& =& \text{6}\text{.0}\text{M}\text{×}\text{0}\text{.0072}\text{L}\\ & =& \text{0}\text{.0432}\text{mol}\end{array}$

Conclusion

The number of moles of $\text{NaOH}$ is 0.0432 mol.

Interpretation Introduction

(b)

Interpretation:

The molarity of resulting solution is to be calculated.

Concept introduction:

Molarity is defined as number of moles of solute dissolved in 1 liter of solution.

It is expressed as,

$\text{Molarity}=\frac{\text{Moles}\text{of}\text{solute}}{\text{Volume}\text{of}\text{solution}\text{in}\text{liter}}$

Answer to Problem 1ASA

The molarity of resulting solution is 0.108 M.

Explanation of Solution

To find the molarity of NaOH solution, (M₂):

Calculate the molarity of NaOH solution using the final volume of NaOH. $\text{Molarity}=\frac{\text{Moles}\text{of}\text{solute}}{\text{Volume}\text{of}\text{solution}\text{in}\text{liter}}$

Substitute the values of moles of NaOH and final volume of NaOH solution (V₂).

$\begin{array}{rll}\text{Molarity}& =& \frac{\text{0}\text{.0432}\text{mol}}{\text{0}\text{.4}\text{L}}\\ & =& \text{0}\text{.108}\text{M}\end{array}$

We also calculate the molarity of NaOH solution as

$\begin{array}{rll}{\text{M}}_{\text{1}}{\text{V}}_{\text{1}}& =& {\text{M}}_{\text{2}}{\text{V}}_{\text{2}}\\ \left(\text{6}\text{.0}\text{M}\right)\text{(7}\text{.2}\text{mL)}& =& {\text{M}}_{\text{2}}\text{(400}\text{.0}\text{mL)}\\ \therefore {\text{M}}_{\text{2}}& =& \text{0}\text{.108}\text{M}\end{array}$

Therefore, the molarity of resulting solution is 0.108 M.

Conclusion

The molarity of resulting solution is 0.108 M.