Question

The charge carried by 1 millimole of ${M^{n + }}$ ions is 193 coulombs. The value of n is :
(A) 1
(B) 2
(C) 3
(D) 4

Answer 1

If the metal has a positive charge of +1, then one mole of the metal ions will have a total charge of 96500 coulombs. One mole of substance contains $6.022 \times {10^{23}}$ particles. We can use this information and calculate the charge on the ion given.

Complete answer:
We are given that ion has a charge of +n.
- We know that charge of one mole of electrons is 96500 coulombs. This was discovered by Faraday. So, this is also used as a Faraday’s constant as well in electrochemistry. Now, here positive charge is supposed to be taken in consideration but this positive charge has same value as the electron's charge.
- So, we can say that one mole of these ${M^{n + }}$ ions will have $n \times 96500$ coulomb charge.
Now, we are asked to find the charge of 1 millimole of these ions.
- We know that one mole of any substance contains $6.022 \times {10^{23}}$ particles. So, we can say that 1 millimole of the ions will contain $6.022 \times {10^{23}} \times {10^{ - 3}} = 6.022 \times {10^{20}}$ ions.
Now, we can write that $6.022 \times {10^{20}}$ ions contain 193 coulombs of charge as given in the question. So, $6.022 \times {10^{23}}$ ${M^{n + }}$ ions contain $n \times 96500$ coulomb of charge.
So, we can write that $n \times 96500 = \dfrac{{6.022 \times {{10}^{23}} \times 193}}{{6.022 \times {{10}^{20}}}}$
Now, we will simplify it.
Thus, $n \times 96500 = 1000 \times 193$
So, $n = \dfrac{{139000}}{{96500}} = 2$
Thus, we can say that n=2.

Therefore, the correct answer is (B).

Note:
In order to solve this problem, we need to remember that the charge on a positively charged ion whose charge is n will have a total charge of $n \times 96500$ coulombs. Do not forget that milli prefix suggests ${10^{ - 3}}$with respect to the original unit.