A metal chloride contains 49.5% chlorine. The | vedclass

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(A) According to Dulong-Petit law,approximate atomic weight $\approx \frac{6.4}{	ext{specific heat}} = \frac{6.4}{0.064} = 100$. In the metal chlorid

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$108.6$

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(A) According to Dulong-Petit law,approximate atomic weight $\approx \frac{6.4}{	ext{specific heat}} = \frac{6.4}{0.064} = 100$. In the metal chloride,mass of chlorine = $49.5 \ g$,so mass of metal = $100 - 49.5 = 50.5 \ g$. Equivalent weight of metal = $\frac{	ext{mass of metal}}{	ext{mass of chlorine}} 	imes 35.5 = \frac{50.5}{49.5} 	imes 35.5 \approx 36.21$. Valency = $\frac{	ext{approximate atomic weight}}{	ext{equivalent weight}} = \frac{100}{36.21} \approx 2.76 \approx 3$. Exact atomic weight = $	ext{equivalent weight} 	imes 	ext{valency} = 36.21 	imes 3 = 108.63$.

$A$ metal chloride contains $49.5\%$ chlorine. The specific heat of the metal is $0.064 \ cal \ g^{-1} \ ^\circ C^{-1}$. What is the exact atomic weight of the metal?

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