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

Answer:

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 -

$M = \frac{n}{V} \times 1000$

Where,

$M$ = molarity of the solution

$n$ = number of moles of Na

$V$ = volume of solution in mL

From the above equation,

$n=\frac{M\times V}{1000}$ = $\frac{0.0350 \times 800}{1000}$ = 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

$\frac{1}{2} \times 0.028$ = 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.