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

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