Question

3.7g of an oxide of a metal was heated with charcoal. The liberated $CO _{2}$ was absorbed in caustic soda solution and weighed $1.0 g$. If the sp. heat of the metal is $0 \cdot 095$, calculate the valency and exact atomic weight of the metal.

Answer 1

SOLUTION : Wt. of oxide of a metal $=3 \cdot 7 g$

Wt. of $CO _{2}$ formed $=1 \cdot 0 g$

Oxygen in $1.0 g CO _{2}$ has been obtained from $3.7 g$ metallic oxide. Hence the weight of oxygen present in metallic oxide is the same as that in $1.0 g CO _{2}$.

Wt. of oxygen in $1 \cdot 0 g CO { }_{2}=\frac{32}{44} g =\frac{8}{11} g$

and, therefore, wt. of metal $=3 \cdot 7-\frac{8}{11}=\frac{40 \cdot 7-8}{11} g =\frac{32 \cdot 7}{11} g$

$\because$ Equivalent weight $=\frac{\text { wt. of metal }}{\text { wt. of oxygen }} \times 8=\frac{32 \cdot 7}{11} \times \frac{11}{8} \times 8=32.7$

By Dulong and Petit's law,

$\text { rough atomic weight }=\frac{6.4}{\text { sp. heat }}=\frac{6.4}{0.095}=67.36$

and valency $=\frac{\text { at. } wt \text {. }}{\text { eq. } wt \text {. }}=\frac{67 \cdot 36}{32 \cdot 7}=2 \cdot 06 \approx 2$.

Exact atomic weight $=32 \cdot 7 \times 2= 6 5 \cdot 4$