Question

Specific heat and equivalent wt. of a metal are $=0.057cal{g}^{-1}$ and $38.24$ respectively. It's exact atomic weight is?

• A. $114.72g$
• B. $38.297g$
• C. $110.9g$
• D. uncertain

Answer 1

Answer: (A)

$114.72g$

Solution:- (A) $114.72g$.

Given that specific heat of metal $=0.057cal{g}^{-1}$

According to Dulong Petit law,

$\text{Approximate atomic weight}=\frac{6.4}{\text{specific heat}}=\frac{6.4}{0.057}=112.28g$

Equivalent weight of metal $=38.24g\left(\text{Given}\right)$

Now applying-

$\text{valency}=\frac{\text{Approximate atomic weight}}{\text{Equivalent weight}}=\frac{112.28}{38.24}=2.93\approx 3$

$\therefore \text{Exact atomic weight}=\text{valency}\times \text{Equivalent weight}$

$\Rightarrow \text{Exact atomic weight}=3\times 38.24=114.72g$

Hence the exact atomic weight of metal is $114.72g$.