Respuesta :
Answer: The molarity of [tex]Cu^{2+}[/tex] ions in the given amount of sample is [tex]4.38\times 10^{-5}M[/tex]
Explanation:
We are given:
Mass of sample = 20 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 20}=0.014g[/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]
We are given:
Mass of solute (copper (II) sulfate) = 0.014 g
Molar mass of copper (II) sulfate = 159.6 g/mol
Volume of solution = 2.0 L
Putting values in above equation, we get:
[tex]\text{Molarity of solution}=\frac{0.014g}{159.6g/mol\times 2.0L}\\\\\text{Molarity of solution}=4.38\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]4.38\times 10^{-5}M[/tex]
Hence, the molarity of [tex]Cu^{2+}[/tex] ions in the given amount of sample is [tex]4.38\times 10^{-5}M[/tex]
