Chapter 19, Problem 57R

To determine

To calculate: The number of moles in the 469 g of $\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ .

Answer to Problem 57R

The number of moles present in the 469 g of $\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ is $2.5\text{moles}$ .

Explanation of Solution

Given info:

The mass of the copper sulphate is 469 g

The molar mass of the $\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ is ${\text{M}}_{\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}}\text{=187}\text{.56 g/mol}$

Formula used:

The molar mass formula is calculated as,

  $\text{Number}\text{of}\text{moles}\text{=}\frac{\text{mass}\text{of}\text{substance}}{\text{molar}\text{mass}}$

Calculation:

By substituting the given values in the above formula, the number of moles can be calculated.

  $\begin{array}{c}\text{Number}\text{of}\text{moles}\text{=}\frac{\text{mass}\text{of}\text{substance}}{\text{molar}\text{mass}}\\ \text{Number}\text{of}\text{moles}=\frac{\text{469 g}}{\text{187}\text{.56 g/mol}}\\ =2.5\text{ mol}\end{array}$

Conclusion:

Thus, the number of moles is $2.5\text{moles}$ .