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?

QUESTION POSTED AT 23/05/2020 - 04:42 AM

Answered by admin AT 23/05/2020 - 04:42 AM

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
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