how much of sodium carbonate would be required to make 250 ml and 200 ml solution of 0.05m
1
Expert's answer
2018-02-10T06:18:07-0500
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.
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