Question

4.215 g of a metallic carbonate was heated in a hard glass tube and the CO₂ evolved was found to measure 1336 ml at 27 and 700 mm pressure. What is the equivalent weight of the metal?

Answer 1

Step 1: Given data :

$$\begin{gathered} P_1=760mm \\ {P}_2=700mm \end{gathered}$$

$$\begin{gathered} V_1=?V_2=1336mL \\ {T}_1=273K,T_2=27+273=300k \end{gathered}$$

Step 2: Formula used

${\mathrm{P}}_1{\mathrm{V}}_1/{\mathrm{T}}_1={\mathrm{P}}_2{\mathrm{V}}_2/{\mathrm{T}}_2$ ----------------------1)

Step 3: Calculation of Volume

From 1)

$$\begin{gathered} 760\times V_1/273=700\times 1336/300 \\ {V}_1=7\times 1336\times 273/3\times 760 \\ =1119.7mL \\ =1.119L \end{gathered}$$

Step 4: Evaluating Weight of Carbon dioxide

Therefore $1.119LC{O}_2=4.215g$ metallic carbonate

$22.4\mathrm{L}{\mathrm{CO}}_2$ is given by $\frac{4.215\times 22.4}{1.119}g$ metallic carbonate

$=84.37g$

Step 5: Calculating Equivalent weight of metal

Metal carbonate is

$$\begin{gathered} {\mathrm{MCO}}_3=\mathrm{M}+12+48 \\ =\mathrm{M}+60 \end{gathered}$$

Atomic weight of Metal

$$\begin{gathered} \mathrm{M}=84.3-60 \\ =24.3 \end{gathered}$$

Equivalent weight of metal =$\frac{Molecularweight}{2}$

$$\begin{gathered} =\frac{24.3}{7} \\ =12.18 \end{gathered}$$