Explanation:
[tex]Molarity=\frac{n}{V(L)}[/tex]
n = moles of compound
V =Volume of the solution in Liters
Moles of potassium sulfate ,n= [tex]\frac{1.80\times 10^{-4} g}{174 g/mol}[/tex]
Volume of the solution = 1.50 L
Molarity of the solution = [tex]M=\frac{1.80\times 10^{-4} g}{174 g/mol\times 1.50 L}[/tex]
Molarity of potassium sulfate solution:
[tex][K_2SO_4]=M=6.8965\times 10^{-7} mol/L[/tex]
1 mole of potassium sulfate gives 2 moles of potassium ions and 1 mole of sulfate ions.
Molarity of potassium ions in the solution:
[tex][K^+]=2\times M=2\times 6.8965\times 10^{-7} mol/L=1.3793\times 10^{-6} mol/L[/tex]
Molarity of sulfate ions in the solution:
[tex][SO_4^{2-}]=1\times 6.8965\times 10^{-7} mol/L=6.8965\times 10^{-7} mol/L[/tex]
Concentration in ppm:
ppm = Milligram of compo present in 1 liter solution.
ppm = Molarity × Molar mass of compound × 1000
Concentration of potassium ion in ppm:
[tex]1.3793\times 10^{-6} mol/L\times 40 g/mol\times 1000[/tex]
[tex][K^+]=0.05517 ppm[/tex]
Concentration of sulfate ions in ppm:
[tex]6.8965\times 10^{-7} mol/L\times 96 g/mol\times 1000[/tex]
[tex][SO_4^{2-}]=0.06620 ppm[/tex]
