Titration chemistry

Question

A 250 mL solution of 2.0 M NaCl is mixed with 350 mL solution of 0.50 M NaCl. What is the concentration of the resulting solution?

Answer 1

Moles of NaCl from 250 mls of 2.0 M NaCl = 250 ml x 1 L / 1000 mls x 2 mol / L = 0.50 mols NaCl

Moles of NaCl from 350 mls of 0.50 M NaCl = 350 ml x 1 L / 1000 mls x 0.50 mol / L = 0.175 mols NaCl

Total moles NaCl = 0.50 + 0.175 = 0.675 mols NaCl

Total volume = 250 mls + 350 mls = 600 mls = 0.600 L

Final concentration of NaCl in the resulting solution = 0.175 mols NaCl / 0.600 L = 0.29 M (2 sig. figs.)