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kkurt [141]
3 years ago
13

The concentration of copper(II) sulfate in one brand of soluble plant fertilizer is 0.070% by weight. If a 21.5 g sample of this

fertilizer is dissolved in 2.0 L of solution, what is the molar concentration of Cu2
Chemistry
1 answer:
Semmy [17]3 years ago
4 0

Answer:

The molar concentration of Cu^{2+} ions in the given amount of sample is 4.73\times 10^{-5}M

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

,=\frac{0.07}{100}\times 21.5}\\=0.0151g of copper (II) sulfate is present.

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}

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

\text{Molarity of solution}=\frac{0.0151g}{159.6g/mol\times 2.0L}\\\\\text{Molarity of solution}=4.73\times 10^{-5}M

The chemical equation for the ionization of copper (II) sulfate follows:

CuSO_4\rightarrow Cu^{2+}+SO_4^{2-}

1 mole of copper (II) sulfate produces 1 mole of copper (II) ions and sulfate ions

Molarity of copper (II) ions = 4.73\times 10^{-5}M

Hence, the molar concentration of Cu^{2+} ions in the given amount of sample is 4.73\times 10^{-5}M

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