Question

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

Answer 1

1.484 grams.

Step-by-step 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 -

`M = (n)/(V) * 1000`

Where,

`M` = molarity of the solution

`n` = number of moles of Na

`V` = volume of solution in mL

From the above equation,

`n=(M* V)/(1000)` =
`(0.0350 * 800)/(1000)` = 0.028 moles.

Now, we know that sodium carbonate dissociates as

`Na_2CO_3` ⇒
`2Na^+ + CO_3^-`

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

`(1)/(2) * 0.028` = 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.