Answer:
a) 0.400 mol/L
b) 9.20 ppt
c) 91.96 ppm
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
[tex]C = \frac{n/ mol}{V / L} \\MW=\frac{m/g}{n/mol} \\C = \frac{m}{MW} .\frac{1}{V} \\\\[/tex]
For Na₂CrO₄:
[tex]C = \frac{3.2473g}{161.97g/mol} .\frac{1}{0.1L} \\C = 0.200 mol/L[/tex]
The molarity of Na will be the double of the molarity of the molarity of Na₂CrO₄:
[tex]C = 0.400 mol/L[/tex]
b) parts per thousand (ppt)
[tex]m ppt = \frac{m}{1000 g} \\m = MW.n\\m ppt = \frac{MW.n}{1000g} \\m ppt = \frac{22.9898.0.04}{1000} \\\\m = 9.20 g\\\\Na: \frac{9.20 g}{1000 g solution} = 9.20 ppt[/tex]
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
[tex]m=MW.n\\m = 22.9898 . 0.004\\m = 0.0916 g\\m = 91.6 mg\\\\Na = 91.96 ppm[/tex]
