The specific heat of a metal is 0.031 cal g | vedclass

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(C) According to the $Dulong$ and $Petit's$ law,the approximate atomic weight is given by: Approximate atomic weight $\approx \frac{6.4}{	ext{Specifi

Vedclass · Accepted Answer

$207.2$

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(C) According to the $Dulong$ and $Petit's$ law,the approximate atomic weight is given by: Approximate atomic weight $\approx \frac{6.4}{	ext{Specific heat}} = \frac{6.4}{0.031} \approx 206.45$. Now,calculate the valency: Valency = $\frac{	ext{Approximate atomic weight}}{	ext{Equivalent weight}} = \frac{206.45}{103.6} \approx 1.99 \approx 2$. Therefore,the exact atomic weight = $	ext{Equivalent weight} 	imes 	ext{Valency} = 103.6 	imes 2 = 207.2$.

The specific heat of a metal is $0.031 \ cal \ g^{-1} \ ^\circ C^{-1}$ and its equivalent weight is $103.6$. What is the exact atomic weight of the metal?

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