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:
$0.065 mol \times 40 = 2.6g$
Mass of solid needed if 98% pure by mass:
$2.6g\times \frac{100}{98} \approx 2.65g$