Question

A 100% pure sample of a divalent metal carbonate weighing 2gm on complete thermal decomposition releases 448cc of carbon dioxide at STP. The equivalent mass of the metal is:
(A) 40
(B) 20
(C) 12
(D) 56

Answer 1

Divalent metal ion stands for the metal which has oxidation number of +2 and will combine with one mole of carbonate ion to form its carbonate. The formula to find the equivalent weight of metal is
${\text{Equivalent weight = }}\dfrac{{{\text{Atomic weight}}}}{{{\text{Oxidation state}}}}$

Complete step by step solution:
We are given that the metal atom is divalent means its valency is +2. So, the formula of that metal carbonate will be$MC{O_3}$. Now, we are given that this upon thermal decomposition produces carbon dioxide gas. This can be expressed in terms of reaction as
$MC{O_3}\xrightarrow{\Delta }MO + C{O_2}$
- It is given to us that upon decomposition of 2gm of the metal carbonate, we get 448cc of carbon dioxide gas.
- Remember that cc stands for $c{m^3}$ and we know that 1$c{m^3} = 1mL$. So, the volume of a given gas is 448mL or 0.448L as 1L=1000mL.
We know that if the volume of the gas is 22.4L at STP, then the amount of gas is 1 mole.
So, we can write that 448mL of the gas is produced from 2gm of carbonate.
Then if 22400mL of carbon dioxide gas is obtained, then mass of the carbonate will be $\dfrac{{2 \times 22400}}{{448}} = 100$gm
Here, we have used 22400mL of carbon dioxide gas because it is the volume of 1 mole of carbon dioxide gas at STP. Thus, we can say from the reaction that the molecular mass of the metal carbonate will be 100$gmmo{l^{-1}}$.
Now, we can write that Molecular weight of $MC{O_3}$ = Atomic weight of metal + Atomic weight of C + 3(Atomic weight of O)
$\therefore 100 = x + 12 + 3(16)$
$\therefore x = 100 - 60 = 40gmmo{l^{-1}}$
Thus, we obtained that the atomic mass of metal is 40 grams per mole.
Now, we now that for metals, ${\text{Equivalent weight = }}\dfrac{{{\text{Atomic weight}}}}{{{\text{Oxidation state}}}}$
${\text{Equivalent weight = }}\dfrac{{40}}{2} = 20gmmo{l^{-1}}$

Thus, the correct answer is (B) 20.

Note: Do not misinterpret that as metal is divalent, it will combine with two moles of carbonate ions, actually carbonate ion has an overall charge of -2 and so a divalent metal ion will always form $MC{O_3}$ type of carbonate.