Question

A zinc rod weighing 25g was kept in 100 mL of a 1 M CuSO4 solution. After certain time the molarity of Cu2+ in the
solution was found to be 0.8. What will be the molarity of Zn in solution?

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Answer 1

Answer:

Explanation:

Zn+CuSO4========>ZnSO4+Cu.

Zinc metal being more reactive and will displace copper from CuSO4 Solution and copper will be deposited as a red precipitate.

The Sulphate ion (SO42–) will not undergo any change and it's concentration will be the same 1M, even after the conversion of 0.7 M of Cu 2+ to Cu.

Answer 2

Given,

weight of  $Zn$ rod = 25g

volume of $CuSO_{4}$ solution = 100 mL

molarity of $CuSO_{4}$ solution = 1 M

molarity of $Cu^{2+}$ in solution = 0.8 M

To find,

The molarity of $Zn$ in the solution

Solution,

No. of moles required by $Cu^{2+}$ = Molarity × Volume

                                                    = (1-0.8) × $\frac{100L}{1000}$

                                                    = 0.02 mole

we know, the molar mass of $Zn$ = 65.4g/mol

so, the mass of $Zn$ that gets oxidized = 0.02 × 65.4g/mol

                                                              = 1.308 g

Therefore, the mass of $Zn$ left = 25 - 1.308 g

                                                  = 23.69 g

                                                    (This will be your final answer.)