Question

shift 2. a. How many grams of solid Na2CO3 is required to prepare 300. mL of 0.500 M Na2CO solution? b. How many mL of the above solution is diluted with water to prepare 300. mL of 0.0250 M Na2CO3 solution?

Answer 1

Mole = Molarity × volume

=> 0.50 ×0.30

= 0.150 mol

Mass = 0.150×106

=> 15.9 g

B) from dilution law,

M1V1 = M2V2

V1×0.50 = 300×0.025

V1 = 15 mL

So we have to add , 300-15=> 285 mL water added.