Question

The equivalent weight of metal whose 1 gram carbonate on heating gives 0.6 gram of metal oxide, is
A. 25 grams
B. 30 grams
C. 35 grams
D. 40 grams

Answer 1

Hint: Write down the above chemical reaction and divide 1 gram of the metal carbonate among the products, MO and $C{O_2}$ formed. Calculate the molecular mass of $C{O_2}$ and $C{O_3}^ - $ and then the equivalent mass of $C{O_2}$, $C{O_3}^ - $ and $MC{O_3}$. The required equivalent weight of metal is the difference in the equivalent weight of $MC{O_3}$ and the equivalent weight of $C{O_3}^ - $.

Complete Step by step answer:
According to the question the chemical reaction is:
$
  MC{O_3} \to MO + C{O_2} \\
  {\text{ }}\left( {{\text{1g}}} \right){\text{ }}\left( {{\text{0}}{\text{.6g}}} \right){\text{ }}\left( {{\text{0}}{\text{.4g}}} \right) \\
 $
Given, the metal carbonate gives 0.6 gram of metal oxide, therefore, \[\left( {1 - {\text{0}}.6g = 0.4g} \right){\text{C}}{{\text{O}}_2}\] is left over.
The molecular mass of \[C{O_2}\] (C – 12 and O - 16) is 44 grams and $C{O_3}^ - $ is 60 grams.
Therefore, the equivalent weight of carbon dioxide is calculated as:
\[
 Equivalent{\text{ }}weight{\text{ }}of{\text{ }}substance\left( {C{O_2}} \right) = \dfrac{{{\text{formula mass of substance}}}}{{{\text{total charge on cationic part}}}} \\
\Rightarrow Equivalent{\text{ }}weight{\text{ }}of{\text{ }}substance\left( {C{O_2}} \right) = \dfrac{{44}}{2} \\
\Rightarrow Equivalent{\text{ }}weight{\text{ }}of{\text{ }}substance\left( {C{O_2}} \right) = 22{\text{ g}} \\
 \]
As, 0.4 gram of $C{O_2}$ is released from 1 gram of metal carbonate. The carbonate which releases 22 grams of $C{O_2}$ is the equivalent mass of the metal carbonate.
So, 22g of $C{O_2}$ will be released from $\dfrac{{1 \times 22}}{{0.4}} = 55g$ of metal carbonate.
Then, the equivalent weight of the carbonate part of metal carbonate is $\dfrac{{60}}{2} = 30g$.
The equivalent weight of metal is the difference in the equivalent weight of $MC{O_3}$ and the equivalent weight of $C{O_3}^ - $.
\[equivalent{\text{ }}weight{\text{ }}of{\text{ }}metal = 55 - 30 = 25g\].
The equivalent weight of metal whose 1 gram carbonate on heating gives 0.6 gram of metal oxide, is 25g.

Hence, the correct option is (A).

Note: Students must always proceed with balanced chemical equations and do not forget to mention the units of the quantities. $MC{O_3}$ comprises metal component M, and non-metal component $C{O_3}^ - $. So, the required equivalent weight of the metal component M, will less than that of the total equivalent weight of the compound $MC{O_3}$