Question

The amount of anhydrous na2co3 present in 250 ml of 0.25 m solution is

Answer 1

Molar mass of Na2CO3 = 46 + 12 + 48 = 106 g/mol

Volume of solution = 250 mL = 0.25L

Molarity = 0.25 M

Molarity, M = moles of solute,n /Volume of solution in L

0.25 = n/0.25

n = 0.25 x 0.25 = 0.0625 mol

Mass of 0.0625 mol of sodium carbonate = 0.0625 x 106 = 6.625 g

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