Chapter 13, Problem 13.23QP

Interpretation Introduction

Interpretation:

The molality and the molarity of ${\text{NH}}_{\text{3}}$ solution is to be determined.

Concept introduction:

Number of moles from its given mass is,

$\text{Number of moles}\text{=}\frac{\text{Given}\text{mass}}{\text{Molecular}\text{mass}}$

Equation for density is,

$\text{Density}\text{=}\frac{\text{Mass}}{\text{Volume}}$

Molality (m):

Molality is the number of moles of solute present in one kilogram of solvent.

$\text{Molality (m) =}\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{1kg}\text{of}\text{solvent}}$

Molarity (M):

Molarity is number of moles of the solute present in the one liter of the solution.

$\text{Molarity (M) =}\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{1}\text{liter}\text{of}\text{solution}}$

Answer to Problem 13.23QP

The molarity of ammonia solution is $\text{18}\text{.28}M$

The molarity of ammonia solution is $\text{2}\text{.69}\text{M}$

Explanation of Solution

Given,

$\begin{array}{c}\text{Mass of solute}\left({\text{NH}}_{\text{3}}\right)\text{=}\text{35}\text{.0}\text{g}\\ \text{Mass of solvent}\left(\text{water}\right)\text{= }\text{75}\text{.0}\text{g}\text{=}0.075kg\\ \text{Density of the}\text{solution}\text{=}\text{0}\text{.982}\text{g/mL}\\ \text{Molecular}\text{mass}{\text{of NH}}_{\text{3}}=17.031g\end{array}$

The molality of ${\text{NH}}_{\text{3}}$ solution is calculated

$\begin{array}{c}\text{Number of moles}\text{=}\frac{\text{Given}\text{mass}}{\text{Molecular}\text{mass}}\\ =\frac{35g}{17.031g}\\ =2.055mol\end{array}$

Therefore,

The molality of ${\text{NH}}_{\text{3}}$ solution is,

$\begin{array}{c}\text{Molality (m) =}\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{1kg}\text{of}\text{solvent}}\\ =\frac{2.055mol}{0.075kg}\\ =27.5m\end{array}$

The molality of ${\text{NH}}_{\text{3}}$ solution is $27.5m$

The molarity of ${\text{NH}}_{\text{3}}$ solution is calculated.

$\text{Density}\text{=}\frac{\text{Mass}}{\text{Volume}}$

$\begin{array}{c}Volume\text{of solvent}\left(\text{water}\right)\text{= }\frac{Mass}{Density}\\ =\frac{75g}{0.982\text{g/mL}}\\ =763.7mL\\ =0.7637L\end{array}$

Therefore,

The molarity of ${\text{NH}}_{\text{3}}$ solution is,

$\begin{array}{c}\text{Molarity (M) =}\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{1}\text{liter}\text{of}\text{solution}}\\ \text{=}\frac{\text{2}\text{.055}\text{mol}}{\text{0}\text{.7637}\text{L}}\\ \text{=}\text{2}\text{.69}\text{M}\end{array}$

The molarity of ${\text{NH}}_{\text{3}}$ solution is $\text{2}\text{.69}\text{M}$

Conclusion

The molarity of ammonia solution was calculated as $\text{18}\text{.28}M$.

The molarity of ammonia solution was calculated as $\text{2}\text{.69}\text{M}$.