Question

how much of sodium carbonate would be required to make 250 ml and 200 ml solution of 0.05m

Answer 1

c = n/V = m/(M*V);
M(Na2CO3) = 106 g/mol;
1. V = 250 mL;
m = c * M * V = 0.05 * 106 * 0.25 = 1.325 g.
2. V = 200 mL;
m = c * M * V = 0.05 * 106 * 0.20 = 1.06 g.