Question

One ampere of current is passed for $9650$ seconds through molten $AlC{l_3}$ . What is the weight in grams of $Al$ deposited at cathode?
(Atomic weight of $Al$ $= 27$ )
(A) 0.9
(B) 9.0
(C) 0.09
(D) 90.0

Answer 1

In order to this question, to calculate the weight of $Al$ deposited at cathode, firstly we will mention the oxidation number of $Al$ . And then we will apply the formula of finding the weight that is related to the current and the time.

Complete answer: Given electric current is, $i = 1A$ ,
Time taken to pass the current, $t = 9650\sec$
As we know that, the oxidation number of $Al$ in $AlC{l_3}$ $= + 3$
Now, we will apply the formula to find the weight: (as current and time is given)
$w = \dfrac{{{E_{eq}}it}}{{96500}} \\\ \Rightarrow w = \dfrac{{27 \times 1 \times 9650}}{{96500}} \\\ = \dfrac{9}{{10}} = 0.9gm \\\$
Therefore, the required weight of $Al$ deposited at cathode is $0.9gm$ .
Hence, the correct option is (A) 0.9 .

Note:
The total of the atomic weights of all the atoms in the chemical formula of a substance (element or compound) is the formula weight. The sum of all the atomic weights of all the atoms in a molecule of a molecular substance is the molecular weight of that substance.