Question

Fe(s) + CuSO4(aq) <===> Cu(s) + FeSO4(aq)

Suppose an industrial quality-control chemist analyzes a sample from a copper processing plant in the following way. He adds powdered iron to a 400.mL copper (II) sulfate sample from the plant until no more copper will precipitate. He then washes, dries, and weighs the precipitate, and finds that it has a mass of 89.mg .

Calculate the original concentration of copper (II) sulfate in the sample. Round your answer to 2 significant digits.

Answer 1

Answer : The original concentration of copper (II) sulfate in the sample is, $5.6\times 10^{-1}g/L$

Explanation :

Molar mass of Cu = 63.5 g/mol

First we have to calculate the number of moles of Cu.

Number of moles of Cu = $\frac{\text{Mass of Cu}}{\text{Molar mass of Cu}}=\frac{89\times 10^{-3}g}{63.5g/mol}=1.40\times 10^{-3}mole$

Now we have to calculate the number of moles of $CuSO_4$

Number of moles of Cu = Number of moles of $CuSO_4$

Number of moles of $CuSO_4$ = $1.40\times 10^{-3}mole$

Now we have to calculate the molarity of $CuSO_4$

$\text{Molarity}=\frac{\text{Moles of }CuSO_4\times 1000}{\text{Volume of solution (in mL)}}$

Now put all the given values in this formula, we get:

$\text{Molarity}=\frac{1.40\times 10^{-3}mole\times 1000}{400.mL}=0.0035M$

To change mol/L into g/L, we need to multiply it with molar mass of $CuSO_4$

Molar mass of $CuSO_4$= 159.609 g/mL

Concentration in g/L = $0.0035M\times 159.609g/mol=0.5586g/L\approx 5.6\times 10^{-1}g/L$

Thus, the original concentration of copper (II) sulfate in the sample is, $5.6\times 10^{-1}g/L$