The specific heat of a metal M is 0.25 J g^ | vedclass

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Question

(C) According to Dulong-Petit's law,the approximate atomic weight is calculated as: Atomic weight $\approx \frac{6.4}{	ext{Specific heat}} = \frac{6.

Vedclass · Accepted Answer

$24$

Vedclass · Answer

(C) According to Dulong-Petit's law,the approximate atomic weight is calculated as: Atomic weight $\approx \frac{6.4}{	ext{Specific heat}} = \frac{6.4}{0.25} = 25.6$. The valency $(n)$ of the metal is calculated as: $n = \frac{	ext{Approximate atomic weight}}{	ext{Equivalent weight}} = \frac{25.6}{12} \approx 2.13$. Since valency must be a whole number,we take $n = 2$. Therefore,the exact atomic weight is: Atomic weight = $	ext{Valency} 	imes 	ext{Equivalent weight} = 2 	imes 12 = 24$.

The specific heat of a metal $M$ is $0.25 \ J \ g^{-1} \ K^{-1}$. Its equivalent weight is $12$. What is its atomic weight?

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