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It's C. To calculate the percentage purity you perform this calculation = 45/50 x 100% = 90%izzahzainab said:http://www.xtremepapers.me/CIE/index.php?dir=Cambridge%20O%20Levels/5070%20-%20Chemistry/&file=5070_s08_qp_1.pdf
check question 12
Remember this fact always: during the electrolytic purification of copper, only copper is deposited at the cathode. The impurity collects at the bottom of the electrolytic cell to form an anode slime. So the gain in mass at the cathode is because of copper being deposited there.izzahzainab said:but they say that the impurity is "below" copper in the reactivity series. that means that the 45 g deposit was of impurity. so the sample must be very impure: should be 5/45 *100 ????
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