Question

A student creates a solution by dissolving some Na2SO4 (molar mass = 142.05 g/mol) in enough water to create 0.500 L of solution. If [Na+ ] = 0.125 M then what mass of Na2SO4 was used?

Answer 1

Answer : The mass of $Na_2SO_4$ is, 4.43 grams.

Explanation :

As we know that, when $Na_2SO_4$ dissolve in water then it dissociates to give 2 mole of sodium ion $Na^+$ and 1 mole of sulfate ion $SO_4^{2-}$

The chemical reaction will be:

$Na_2SO_4\rightarrow 2Na^++SO_4^{2-}$

Thus, the concentration of $Na_2SO_4$ = $\frac{\text{Concentration of }Na^+}{2}=\frac{0.125M}{2}=0.0625M$

First we have to calculate the moles of $Na_2SO_4$

$\text{Concentration of }Na_2SO_4=\frac{\text{Moles of }Na_2SO_4}{\text{Volume of solution}}$

$0.0625M=\frac{\text{Moles of }Na_2SO_4}{0.500L}$

$\text{Moles of }Na_2SO_4=0.0312mol$

Now we have to calculate the mass of $Na_2SO_4$

$\text{Mass of }Na_2SO_4=\text{Moles of }Na_2SO_4\times \text{Molar mass of }Na_2SO_4$

Molar mass of $Na_2SO_4$ = 142 g/mol

$\text{Mass of }Na_2SO_4=0.0312mol\times 142g/mol=4.43g$

Thus, the mass of $Na_2SO_4$ is, 4.43 grams.