Question

Calculate the molarity of a solution that contains 3.00 grams Na2SO4 in 25 mL of solution.

Answer 1

Answer:

M Na2SO4 sln = 0.8448 M

Explanation:

• molarity (M) [=] mol/L

∴ mass Na2SO4 = 3.00 g

∴ volume soln = 25 mL = 0.025 L

∴ molar mass Na2SO4 = 142.04 g/mol

⇒ mol Na2SO4 = (3.00 g)*(mol/142.04 g) = 0.02112 mol

⇒ M Na2SO4 sln = (0.02112 mol/0.025 L ) = 0.8448 M