Question

What mass of NaOH is required to prepare 500.00 mL of 0.1300 M NaOH stock solution, if the solid NaOH is only 98.00% pure?

Answer 1

I would say it would depend on what the impurity is. But if the purity is determined as 98% pure by mass, then we can calculate:

Moles of NaOH in the stock solution:

`0.1300M * 500mL = 0.065 mol`
Mass of pure NaOH solid (
`M_(r)=40g/mol`) needed:

`0.065 mol * 40 = 2.6g`
Mass of solid needed if 98% pure by mass:

`2.6g* (100)/(98) \approx 2.65g`