Question

How many milliliters of an aqueous solution of 0.227 M sodium carbonate is needed to obtain 3.55 grams of the salt?

Answer 1

147.6 mL .

Step-by-step explanation:

sodium carbonate , Na₂CO₃

Molecular weight = 106

3.55 gm of sodium carbonate = 3.55 / 106

= .0335 moles

Let the volume of litre required = V

V litre of .227 M solution will contain

V x .227 moles of sodium carbonate . So

V x .227 = .0335

V = .1476 L

= 147.6 mL .