A metal M (specific heat 0.16 J g^{-1} K^{ | vedclass

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Question

(B) According to the Dulong-Petit law, the approximate atomic mass of the metal is $\approx \frac{6.4}{	ext{Specific heat}} = \frac{6.4}{0.16} = 40 \

Vedclass · Accepted Answer

$MCl_2$

Vedclass · Answer

(B) According to the Dulong-Petit law, the approximate atomic mass of the metal is $\approx \frac{6.4}{	ext{Specific heat}} = \frac{6.4}{0.16} = 40 \ g \ mol^{-1}$. In the metal chloride, the percentage of chlorine is $65\%$, so the percentage of metal $M$ is $35\%$. The equivalent weight of the metal is calculated as $\frac{	ext{Mass of metal}}{	ext{Mass of chlorine}} 	imes 35.5 = \frac{35}{65} 	imes 35.5 \approx 19.11$. The valency $n$ is given by $\frac{	ext{Approximate atomic mass}}{	ext{Equivalent weight}} = \frac{40}{19.11} \approx 2.09 \approx 2$. Thus, the formula of the metal chloride is $MCl_2$.

$A$ metal $M$ (specific heat $0.16 \ J \ g^{-1} \ K^{-1}$) forms a metal chloride with $65\%$ chlorine present in it. The formula of the metal chloride will be:

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