(a)A zinc rod weighing 25.00 g was kept in 100 mL of 1 M CuSO4 solution.After a certain time molarity of Cu2+ in solution was 0.8 M.What was the molarity of the sulphate ion(SO42-)?What was the weight of the zinc rod after cleaning?
(b)If the above experiment was done with a copper rod of weight 25g and 50mL of 2M zinc sulphate(ZnSO4)solution,what would be the molarity of Zn2+at the end of the same interval?
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2 Answers
262a) there will be no change in conc. of sulphate ions, hence 1M.
moles of zinc ions formed = moles of copper deposited
= (0.2 M)* (0.1 L) = 0.02 moles = 0.02 * 65.5 =1.31 g of zinc
thus weight of zinc rod = 25.00 - 1.31 = 23.69g
b) in second case Cu wont replace Zn ions (i hope you know this) . hence no change in conc.
