Question

3.2473 g of Na2CrO4 (MW = 161.97 g/mol) is dissolved in 100.0 mL of water. Assuming the solution has a density of 1.00 g/mL, what is the concentration of Na (MW = 22.9898 g/mol) in the solution in units of (a) molarity (M)? (b) parts per thousand (ppt)? (c) 10.0 mL of the solution is then diluted to a final volume of 1000.0 mL. What is the concentration of Na in the diluted solution in units of parts per million (ppm)?

Answer 1

The concentration of Na in the solution is 0.401 M (molarity), 8.943 ppt (parts per thousand), and after dilution it becomes 92.25 ppm (parts per million).

Step-by-step explanation:

To calculate the concentration of sodium (Na) in a solution where 3.2473 g of Na2CrO4 is dissolved in 100.0 mL of water, you need to follow these steps:

1. Determine the moles of Na2CrO4 by dividing the mass of Na2CrO4 by its molar mass:

Moles of Na2CrO4 = 3.2473 g / 161.97 g/mol = 0.02005 mol

1. Calculate the total moles of Na, noting that each mole of Na2CrO4 contains 2 moles of Na:

Moles of Na = 0.02005 mol Na2CrO4 × 2 mol Na/mol Na2CrO4 = 0.0401 mol

1. Convert the volume of the solution to liters and calculate the molarity (M):

Molarity = Moles of Na / Volume (L) = 0.0401 mol / 0.100 L = 0.401 M

1. To find the concentration in parts per thousand (ppt), we first convert grams to milligrams:

Concentration of Na = (Mass of Na / Mass of solution) × 1000 ppt = (0.0401 mol × 22.9898 g/mol) / (100.0 mL × 1.00 g/mL) × 1000 ppt = 8.943 ppt

1. For the dilution part, use the dilution formula M1V1 = M2V2. After dilution the molarity of Na becomes:

M2 = (M1 × V1) / V2 = (0.401 M × 0.0100 L) / 1.000 L = 0.00401 M

1. Finally, calculate the concentration in parts per million (ppm) after dilution:

Concentration in ppm = (Moles of Na after dilution / Volume of solution) × Molar mass of Na × 10¶ = (0.00401 M × 1 L) × 22.9898 g/mol × 10¶ ppm/g = 92.25 ppm

Answer 2

a) 0.400 mol/L

b) 9.20 ppt

c) 91.96 ppm

Step-by-step explanation:

The first step is considereing the dissociation of the salt in water:

Na₂CrO₄ (s) → 2Na⁺ (aq) + CrO₄²⁻ (aq)

This means that for every mol of Na₂CrO₄ 2 mol of Na⁺ will be produced.

a) molarity

`C = (n/ mol)/(V / L) \\MW=(m/g)/(n/mol) \\C = (m)/(MW) .(1)/(V) \\\\`

For Na₂CrO₄:

`C = (3.2473g)/(161.97g/mol) .(1)/(0.1L) \\C = 0.200 mol/L`

The molarity of Na will be the double of the molarity of the molarity of Na₂CrO₄:

`C = 0.400 mol/L`

b) parts per thousand (ppt)

`m ppt = (m)/(1000 g) \\m = MW.n\\m ppt = (MW.n)/(1000g) \\m ppt = (22.9898.0.04)/(1000) \\\\m = 9.20 g\\\\Na: (9.20 g)/(1000 g solution) = 9.20 ppt`

c) parts per million (ppm)

0.400 mol Na ____ 1000 mL solution

x ____ 10 mL solution

x = 0.004 mol Na

The new solution has a concentration of Na of 0.004 mol/L.

ppm = mg/L

`m=MW.n\\m = 22.9898 . 0.004\\m = 0.0916 g\\m = 91.6 mg\\\\Na = 91.96 ppm`