Answer:
4.02g of Na2CO3⸱2H2O must be added completing the volume of the solution to 100mL
Explanation:
A 3%(w/v) solution contains 3g of solute (In this case, Na2CO3) in 100mL of solution.
Assuming we require 100mL of solution we must add 3g of Na2CO3. The reactant that is available is its dihydrate, with molar mass:
106g/mol + 2*MW H2O
106g/mol + 2*18g/mol = 142g/mol
That means the mass of Na2CO3.2H2O that must be added to prepare the solution is:
3g Na2CO3 * (142g/mol Na2CO3.2H2O / 106g/mol Na2CO3) =
<h3>4.02g of Na2CO3⸱2H2O must be added completing the volume of the solution to 100mL</h3>
