Answer:
The molar concentration of [tex]Cu^{2+}[/tex] ions in the given amount of sample is [tex]4.73\times 10^{-5}M[/tex]
Explanation:
Given that,
Mass of sample = 21.5 g
0.07 % (m/m) of copper (II) sulfate in plant fertilizer
This means that, in 100 g of plant fertilizer, 0.07 g of copper (II) sulfate is present
So, in 20 g of plant fertilizer
,[tex]=\frac{0.07}{100}\times 21.5}\\=0.0151g[/tex] of copper (II) sulfate is present.
To calculate the molarity of solution, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}[/tex]
Mass of solute (copper (II) sulfate) = 0.0151 g
Molar mass of copper (II) sulfate = 159.6 g/mol
Volume of solution = 2.0 L
[tex]\text{Molarity of solution}=\frac{0.0151g}{159.6g/mol\times 2.0L}\\\\\text{Molarity of solution}=4.73\times 10^{-5}M[/tex]
The chemical equation for the ionization of copper (II) sulfate follows:
[tex]CuSO_4\rightarrow Cu^{2+}+SO_4^{2-}[/tex]
1 mole of copper (II) sulfate produces 1 mole of copper (II) ions and sulfate ions
Molarity of copper (II) ions = [tex]4.73\times 10^{-5}M[/tex]
Hence, the molar concentration of [tex]Cu^{2+}[/tex] ions in the given amount of sample is [tex]4.73\times 10^{-5}M[/tex]
