Question

4.2g of metallic Carbonate MCO3
was heated in a hard glass tube
and Co2 evolved was found to have
1120ML of volume at STP. The Equivalent weight
of metal is
a)12 b ) 24 c) 18 d) 15​

Answer 1

Answer:

The correct answer is option (a).

Explanation:

$MCO_3\rightarrow MO+CO_2$

Volume of carbon dioxide produced at STP = 1120 mL = 1.12 L

At STP, 1 mol of gas occupies 22.4 L of volume.

Then 1.12 L of volume will be occupied by:

$\frac{1}{22.4}\times 1.12 mol=0.05 mol$

0.05 moles of carbon dioxide are produced.

According to reaction, 1 mole of carbon dioxide are produced from 1 mole of metallic carbonate.

Then. 0.05 mole of carbon dioxide will be produced from:

$\frac{1}{1}\times 0.05 mol=0.05 mol$ of metallic carbonate.

Molar mass of metallic carbonate =m+12 g/mol+3 ×16 g/mol= (m+ 60)g/mol

here , m = atomic mass of metal

Moles of metallic carbonate:

$0.05 mol=\frac{4.2 g}{\text{m+60}}$

m = 24 g/mol

From the formula of metallic carbonate we can see that valency is 2.

$M^{2+}+CO_3^{2-}\rightarrow MCO_3$

The Equivalent weight  of metal is:

$\frac{\text{Atomic mass}}{Valency}=\frac{24 g/mol}{2 eq/mol}=12 g/eq$

Hence, the correct answer is option (a).