Respuesta :
Answer:
2.12 grams of primary-standard-grade sodium carbonate are necessary to prepare 800.0 mL of this solution.
Explanation:
Molarity of sodium ions = [tex][Na^+][/tex] = 0.0500 M
Moles of sodium ions = n
Volume of the solution = V = 800.0 mL = 0.800 L
[tex]Molarity=\frac{n}{V(L)}[/tex]
[tex][Na^+]=\frac{n}{V}[/tex]
[tex]n=[Na^+]\times V=0.0500 M\times 0.800 L=0.04 mol[/tex]
[tex]Na_2SO_4(aq)\rightarrow 2Na^+(aq)+CO_3^{2-}(aq)[/tex]
1 mole sodium carbonates gives 2 moles of sodium ion and 1 mole of carbonate ions.
Then 0.04 moles of sodium ions will be obtained from:
[tex]\frac{1}{2}\times 0.04 mol=0.02 mol[/tex] of sodium m carbonation.
Mass of 0.02 moles of sodium carbonate = 0.02 mol × 106 g/mol= 2.12 g
2.12 grams of primary-standard-grade sodium carbonate are necessary to prepare 800.0 mL of this solution.
