Question

If 8.97 grams of (nh4)2co3 is used to make a 0.250-molar solution, what is the final volume of the solution?

Answer 1

0.3733 L or 373 mL

Solution:

Data Given:

Mass = 8.97 g

M.Mass of (NH₄)₂CO₃ = 96.09 g.mol⁻¹

Molarity = 0.250 mol.L⁻¹

Step 1: Calculate Moles of Ammonium Carbonate as,

Moles = Mass ÷ M.Mass

Putting values,

Moles = 8.97 g ÷ 96.09 g.mol⁻¹

Moles = 0.0933 mol

Step 2: Calculate Volume of Solution as,

Molarity = Moles ÷ Volume

Solving for Volume,

Volume = Mole ÷ Molarity

Putting values,

Volume = 0.0933 mol ÷ 0.250 mol.L⁻¹

Volume = 0.3733 L

Or,

Volume = 373 mL