Question

zn rod is placed in 100 ml of 1M cuso4 solution so that molarity of cu2+ changes to 0.7M.the molarity of so4^2- at this stage will be

Answer 1

When Zinc, Zn rod is placed in 100 ml of 1 M copper (II) sulphate, CuSO4 solution so that molarity of Cu2+ changes to 0.7 M. Then the molarity of S0²⁻₄ at this stage will be calculated as follows;

moles of copper (II) sulphate is given as = molarity x volume/1000

                                     1 x 100/1000 = 0.1 M

Ten the initial concentration is 0.1 x 1000/100

                                                  = 1 M

Hence the molarity of S0²⁻₄ at this stage = 1 - 0.7

                                                                   = 0.3 M

Answer 2

Answer: 1 M SO4^(2-)

Explanation:

Zinc being more reactive reacts with Copper(II)sulphate and Copper precipitates. But sulphate doesn't change and remains the same. As a result molarity will remain at 1M even after reaction

The reaction :

Zn + CuSO4 -> ZnSO4 + Cu (ppt.)