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melamori03 [73]
3 years ago
10

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

Chemistry
1 answer:
Ludmilka [50]3 years ago
3 0

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

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