Question

A 5.0067 g sample of Na2CrO4 (MW = 161.97 g/mol) is dissolved in 250.0 mL of water. Assume the solution has a density of 1.00 g/mL.

A 50.0 mL aliquot of the solution is then diluted to a final volume of 250.0 mL. What is the concentration of Na+ in the diluted solution in units of parts per million (ppm)?

Answer 1

First we need to calculate [Na₂CrO₄]

We have, Molarity = No. of moles of solute / Volume of solution in L

Here No. of moles of Na₂CrO₄ = Mass / MW = 5.0067 g / 161.97 g/mol = 0.030911 mol

Therefore, [Na₂CrO₄] = 0.030911 mol / 0.250 L = 0.1236 M

Now, by using dilution formula we can calculate concentration of dilute solution.

We have dilution formula : M stock x V stock = M dilute x V dilute

Therefore, M dilute = M stock x V stock / V dilute = 0.1236 M x 50 ml / 250 ml = 0.02472 M

One molecule of Na₂CrO₄ contain two Na ⁺ ions, Hence [Na⁺] = 2 x [Na₂CrO₄] = 2 x 0.02472 = 0.04944 M

No. of moles of Na ⁺ = Molarity x volume of solution in L = 0.04944 mol / L x 0.250 L = 0.01236 mol

Mass of Na ⁺ = No. of moles x Molar Mass = 0.01236 mol x 22.99 g / mol = 0.2842 g

i e 250 g solution contain 0.2842 g Na ⁺. ( density = 1 gm / ml )

We have, ppm = (mass of solute / Mass of solution ) x 10 ⁶

( 0.2842 g / 250 g) x 10 ⁶ = 1137 ppm

ANSWER : Concentration of diluted Na ⁺ solution = 1137 ppm