Question

A solution is to be prepared with a sodium ion concentrationof

0.148 mol/L. What mass of sodium sulfate (g) is needed toprepare
2.29 liters of such a solution?

Answer 1

Answer : The mass of sodium sulfate needed is 5.7085 grams.

Explanation : Given,

Concentration of sodium ion = 0.148 mol/L

Volume of solution = 2.29 L

Molar mass of sodium sulfate = 142 g/mole

First we have to determine the moles of sodium ion.

`\text{Concentration of sodium ion}=\frac{\text{Moles of sodium ion}}{\text{Volume of solution}}`

`0.184mol/L=\frac{\text{Moles of sodium ion}}{2.29L}`

`\text{Moles of sodium ion}=0.08035mole`

Now we have to calculate the moles of sodium sulfate.

The balanced chemical reaction will be,

`Na_2SO_4\rightarrow 2Na^++SO_4^(2-)`

As, 2 moles of sodium ion produced from 1 moles of
`Na_2SO_4`

So, 0.08035 moles of sodium ion produced from
`(0.08035)/(2)=0.040175` moles of
`Na_2SO_4`

Now we have to calculate the mass of sodium sulfate.

`\text{Mass of }Na_2SO_4=\text{Moles of }Na_2SO_4* \text{Molar mass of }Na_2SO_4`

`\text{Mass of }Na_2SO_4=0.040175mole* 142g/mole=5.7085g`

Therefore, the mass of sodium sulfate needed is 5.7085 grams.