Chapter 16.2, Problem 12SP

Interpretation Introduction

Interpretation: To calculate the number of moles of ammonium nitrate present in $335ml$ of $0.425M$

  ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$

Concept introduction: Molarity is defined as the ratio of moles of solute to that of the volume of solvent that is used to dissolve the given solute.

Answer to Problem 12SP

The number of moles of ammonium nitrate present in $335ml$ of $0.425M$

  ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$ is $0.142\text{ moles}$ .

Explanation of Solution

Molarity: Molarity is defined as the ratio of moles of solute to that of the volume of solvent which is used to dissolve the given solute.

Molarity is calculated as follows:

  $\text{M=}\frac{\text{n (solute)}}{\text{V (solution in L)}}$

Where,

• $M$ is the molarity that is to be calculated for the particular solution
• $n$ is the number of moles of solute.
• $V$ is the volume of the solution.

Ammonium nitrate dissociates as follows:

  ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}\text{ (aq)}\to {\text{ NH}}_{\text{4}}^{\text{+}}{\text{ (aq) +NO}}_{\text{3}}^{\text{-}}\text{ (aq)}$

The dissociated solution does not contain any ammonium nitrate, so this suggests the concentration to be zero.

Molarity of the solution $=0.425mol{L}^{-1}$

The volume of the solution $\text{=335 mL}$

So, the molarity $\text{=}\frac{\text{n (solute)}}{\text{V (solution in L)}}$

  $0.425=\frac{n(\text{solute})}{335\times {10}^{-3}}$

  $n(\text{solute})=0.425mol{L}^{-1}\times (335\times {10}^{-3}L)$

Moles of the solute $=0.142moles$

Conclusion

The number of moles of Ammonium nitrate present in $335ml$ of $0.425M$

  ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$ is $0.142\text{ moles}$ .