Astandardized solution that is 0.0500 0.0500 m in na + na+ is necessary for a flame photometric determination of the element. how many grams of primary-standard-grade sodium carbonate are necessary to prepare 800.0 800.0 ml of this solution?

2.12 grams of primary-standard-grade sodium carbonate are necessary to prepare 800.0 mL of this solution.

Explanation:

Molarity of sodium ions = = 0.0500 M

Moles of sodium ions = n

Volume of the solution = V = 800.0 mL = 0.800 L

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:

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.

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