200mL of 4.98M of sodium chloride solution is added to an additional 532 ml of water what is the final molarity?

M₂ = 1.9 M

Explanation:

Given data;

Volume of sodium chloride = 200 mL

Molarity of sodium chloride = 4.98 M

Volume of water = 532 mL

Final Molarity = ?

Solution:

M₁V₁ = M₂V₂

M₂ = M₁V₁ / V₂

M₂ = 4.98 M * 200 mL / 532 mL

M₂ = 996 mL. M / 532 mL

M₂ = 1.9 M