How many grams of na2co3 would be needed to prepare 500ml of a 4.00m solution of na2co3?

212 g Na₂CO₃

We are given;

We are required to calculate the mass of Na₂CO₃.

We need to know how molarity is calculated.

Moles = Molarity × Volume

Therefore;

Moles of Na₂CO₃ = 4.00 M × 0.5 L

                             = 2 moles

Thus, moles of Na₂CO₃ is 2 moles

We need to know the relationship between the mass of a compound, molar mass and the number of moles.

Moles = Mass ÷ molar mass

Rearranging the formula;

Mass = Moles × molar mass

Molar mass of Na₂CO₃ is 106 g/mol

Therefore;

Mass of Na₂CO₃ = 2 moles × 106 g/mol

                          = 212 g

Therefore, the mass of Na₂CO₃ that would be needed is 212 g

- lead + phosphoric acid = hydrogen + lead(ii) phosphate