Chapter 10, Problem 86A

Interpretation Introduction

Interpretation: The total mass of a mixture needs to be determined.

Concept Introduction: According to Avogadro’s law, in 1 mol of any substance there are $6.022\times {10}^{23}$ units of that substance. Also, the number of moles of a substance can be calculated as follows:

  $n=\frac{m}{M}$

Here, m is mass and M is the molar mass of the substance.

Explanation of Solution

The given composition of a mixture is $3.50\times {10}^{22}{\text{ formula units Na}}_{\text{2}}{\text{SO}}_{\text{4}}$ , 0.500 mol ${\text{H}}_{\text{2}}\text{O}$ and 7.23 g of AgCl.

Considering each component one by one and converting the given value in grams to calculate the total mass.

  $3.50\times {10}^{22}{\text{ formula units Na}}_{\text{2}}{\text{SO}}_{\text{4}}$

Now, the number of formula units can be converted into the number of moles as follows:

  $\begin{array}{c}n=\frac{1\text{ mol}}{6.022\times {10}^{23}\text{ fromula units}}\times 3.50\times {10}^{22}\text{ formula units}\\ =0.058\text{ mol}\end{array}$

Number of moles of ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}$ is 0.058 mol. Molar mass of ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}$ is 142.04 g/mol; thus, the mass of ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}$ can be calculated as follows:

  $m=n\times M$

Putting the values,

  $\begin{array}{c}m=\left(0.058\text{ mol}\right)\left(142.04\text{ g/mol}\right)\\ =8.24\text{ g}\end{array}$

Thus, the mass of $3.50\times {10}^{22}{\text{ formula units Na}}_{\text{2}}{\text{SO}}_{\text{4}}$ is 8.24 g.

Now, 0.500 mol ${\text{H}}_{\text{2}}\text{O}$ .

Molar mass of ${\text{H}}_{\text{2}}\text{O}$ is 18 g/mol; thus, mass can be calculated as follows:

  $m=n\times M$

Substitute the values,

  $\begin{array}{c}m=\left(0.5\text{ mol}\right)\left(18\text{ g/mol}\right)\\ =\text{9 g}\end{array}$

Thus, the mass of 0.500 mol ${\text{H}}_{\text{2}}\text{O}$ is 9 g.

Also, 7.23 g of AgCl is given. The total mass of the mixture can be calculated as follows:

  $\begin{array}{c}\text{Total mass}=8.24\text{ g}+9\text{ g}+\text{7}\text{.23 g}\\ =\text{24}\text{.47 g}\end{array}$

Therefore, the total mass of the mixture is $\text{24}\text{.47 g}$ .