Answer to Question #174954 in General Chemistry for Liza

Question

How much 2.0 M NaOH stock solution is needed to prepare 300. mL of a 0.050 M NaOH solution?

Answer 1

300/1000 = 0.3 L

Moles of solute = molarity × Litres of solution

= (0.050/ 1L) × 0.3 = 0.015 mol

Molar Mass = 40.00g/mol

= 0.015 × 40.0g/ 1mol = 3.60g

= 3.60 × 2 = 7.2g of 2.0M NaOH