Question

The carbonate of metal is isomorphous with ${\mathrm{MgCO}}_3$ and contains $6.091\%$ of carbon. The atomic weight of the metal is :

• A. 48
• B. 68
• C. 120
• D. 137

Answer 1

The correct option is D

137

Explanation for correct answer:-

Option (D) 137:

Step 1: GIven data

The percentage composition of carbon is $6.091\%$

The carbonate of metal is isomorphous with ${\mathrm{MgCO}}_3$

Step 2: Molecular mass of the compound

We know that the atomic mass of Carbon is 12 and Oxygen is 16

Let us assume the mass of the metal be M

Thus the molecular formula becomes ${\mathrm{MCO}}_3$

$\therefore$The molecular mass of the compound now is

$$\begin{gathered} \mathrm{M}+12+(16\times 3) \\ =\mathrm{M}+12+48 \\ =\mathrm{M}+60 \end{gathered}$$

Step 3: Mass of metal

It is given to us that the percentage of carbon is $6.091\%$

$$\begin{gathered} \therefore \frac{6.091}{100}\times (\mathrm{M}+60)=12 \\ \Rightarrow (\mathrm{M}+60)\times 6.091=1200 \\ \Rightarrow \mathrm{M}+60=\frac{1200}{6.091} \\ \Rightarrow \mathrm{M}+60=197.01 \\ \Rightarrow \mathrm{M}=197-60 \\ \Rightarrow \mathrm{M}=137 \end{gathered}$$

Step 4: Identifying the metal

Now the element with atomic mass $137\mathrm{amu}$ is Barium.

Thus the formula of the compound now becomes ${\mathrm{BaCO}}_3$

Therefore, the atomic weight of the metal is an option (D) $137$ (Barium).