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