Question

The carbonate of a metal contains 50% of the metal, the equivalent mass of the metal atom would be:
A.15
B.30
C.50
D.25

Answer 1

Assume there is 100g of the metal carbonate present. Then find the equivalent weight of the metal using the percentages of the metal, carbonate present and equivalent weight of the carbonate.

Complete answer:
Let us find out what equivalent weight is. Equivalent weight can be defined as the mass of 1 equivalent, rather, the mass of a given substance which is suitable to combine with , in fact displace a finite amount of another substance. The combining weight of part is that the mass which mixes with or displaces 1.008 gram of hydrogen or 8.0 grams of oxygen or 35.5 grams of chlorine. Now, in order to find the equivalent weight of metal in the above metal carbonate, let us assume that 100 g of the metal carbonate is present.
Now, 50% of the metal is present and 50% of carbonate is present. So, in 100g, 50g metal and 50g carbonate is the composition. Let us find the equivalent weight of the carbonate. Equivalent weight is given by:
$Eq.wt=\dfrac{molar\,mass}{valency}$
So, for carbonate, it is
$\begin{aligned} & \dfrac{12+(3\times 16)}{2} \\\ & =\dfrac{60}{2} \\\ & =30 \\\ \end{aligned}$, as carbonate has a charge of 2
Now, let us calculate the equivalent weight of the metal in the compound, which is given by:
$eq.wt/ of /metal=\dfrac{percentage/ of/ metal}{percentage /of /carbonate}\times eq./wt /of /carbonate$
Substituting the values, we get
$\begin{aligned} & \dfrac{50}{50}\times 30 \\\ & =30 \\\ \end{aligned}$

So, the equivalent weight of the metal in the metal carbonate is 30, which gives the correct answer as option B.

Note:
As you can see from the formula, the equivalent weight of the metal depends on both the composition of the substances, as well as the equivalent weight of the carbonate or some other group like oxide attached to it. That means, the equivalent weight of the metal will increase, if the percentage of the metal is present in more quantity, and vice versa.