Chapter U4, Problem 9RE

Interpretation Introduction

Interpretation:The mass of solute in grams that are dissolved in given sample needs to be calculated.

Concept Introduction:The molar concentration of a solute in a solution can be expressed in terms of molarity as moles of solute present in per liter of solution. Mathematically,

  $\text{Molarity = }\frac{\text{number of moles of solute}}{\text{Volume of solution (L)}}$ - (1)

Answer to Problem 9RE

The mass of solute is $\text{35}\text{.51 g}$ .

Explanation of Solution

Given:

500.0 mL sample of 0.50 M sodium sulfate.

Converting the given volume of sample from mL to L as:

  $500\text{ mL}\times \frac{1\text{ L}}{1000\text{ mL}}=\text{ 0}\text{.5 L}$

The formula to calculate number of moles is:

  $\text{number of moles = }\frac{\text{mass of the substance}}{\text{Molar mass of the substance}}$

Now, equation (1) an be rewritten as:

  $\text{Molarity = }\frac{\frac{\text{mass of the solute}}{\text{Molar mass of the solute}}}{\text{Volume of solution (L)}}$ - (2)

The molar mass of sodium sulfate is $\text{142}\text{.04 g/mol}$ .

Substituting the given values in equation (2) to determine the mass of solute as:

  $\begin{array}{l}\text{0}\text{.5 mol/L = }\frac{\frac{\text{mass of the solute}}{\text{142}\text{.04 g/mol}}}{\text{0}\text{.5 L}}\\ \text{0}\text{.5 mol/L = }\frac{\text{mass of the solute}}{\text{0}\text{.5 L}\times \text{142}\text{.04 g/mol}}\\ \text{mass of the solute = 0}\text{.5 mol/L}\times \text{0}\text{.5 L}\times \text{142}\text{.04 g/mol}\\ \text{mass of the solute = 35}\text{.51 g}\end{array}$

Hence, the mass of solute is $\text{35}\text{.51 g}$ .