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}=(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* \text{Molar mass of }Na_2SO_4`

Molar mass of
`Na_2SO_4` = 142 g/mol

`\text{Mass of }Na_2SO_4=0.0312mol* 142g/mol=4.43g`

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