A standardized solution that is 0.0350 M in 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 mL of this solution

1.484 grams.

Explanation:

At first we have to find the number of moles in 0.0350 M of 800.0 mL of Na solution. For this we will use the molarity equation as follows -

Where,

= molarity of the solution

= number of moles of Na

= volume of solution in mL

From the above equation,

= = 0.028 moles.

Now, we know that sodium carbonate dissociates as

⇒

That means 1 mole of sodium carbonate produces 2 moles of Na⁺ ion

So, the moles of sodium carbonate required for 0.028 mole of Na⁺ ion will be

= 0.014 moles.

So, the weight of sodium carbonate required = number of moles × molecular mass

= 0.014 × 105.9888 = 1.484 g

The weight of sodium carbonate required for preparing the solution mentioned in the question is 1.484 grams.

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