Question

Sodium sulfate dissolves as follows: Na2SO4(s) → 2Na+(aq) + SO42- (aq). How many moles of Na2SO4 are required to make 1.0 L of solution in which the Na concentration is 0.10 M?

Answer 1

Answer: The number of moles of $Na_2SO_4$ is 0.05 moles.

Explanation:

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

We are given:

Molarity of solution = 0.10 mol/L

Volume of solution = 1 L

Putting values in above equation, we get:

$0.10mol/L=\frac{\text{Moles of sodium}}{1.0L}\\\\\text{Moles of sodium}=0.10mol$

The chemical reaction for the ionization of sodium sulfate follows the equation:

$Na_2SO_4(s)\rightarrow 2Na^+(aq.)+SO_4^{2-}(aq.)$

By Stoichiometry of the reaction:

2 moles of sodium ions are produced by 1 mole of sodium sulfate

So, 0.10 moles of sodium ions will be produced by = $\frac{1}{2}\times 0.1=0.05moles$ of sodium sulfate.

Hence, the number of moles of $Na_2SO_4$ is 0.05 moles.