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A transition metal $M$ forms a volatile chloride which has a vapour density of $94.8$. If it contains $74.75 \%$ of chlorine the formula of the metal chloride will be
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Verified Answer
The correct answer is:
$\mathrm{MCl}_4$
$\mathrm{MCl}_4$
$74.75 \%$ of chlorine means $74.75 \mathrm{~g}$ chlorine is present in $100 \mathrm{~g}$ of metal chloride.
Weight of metal $=100 \mathrm{~g}-74.75 \mathrm{~g}$ $=25.25 \mathrm{~g}$
Equivalent weight
$$
\begin{aligned}
& =\frac{\text { weight of metal }}{\text { weight of chlorine }} \times 35.5 \\
& =\frac{25.25}{74.75} \times 35.5=12
\end{aligned}
$$
Valency of metal
$$
\begin{array}{r}
=\frac{2 \times \text { V.D. }}{\text { Equivalent wt. of metal }+35.5} \\
=\frac{2 \times 94.8}{12+35.5}=4
\end{array}
$$
$\therefore$ Formula of compound $=\mathrm{MCl}_4$
Weight of metal $=100 \mathrm{~g}-74.75 \mathrm{~g}$ $=25.25 \mathrm{~g}$
Equivalent weight
$$
\begin{aligned}
& =\frac{\text { weight of metal }}{\text { weight of chlorine }} \times 35.5 \\
& =\frac{25.25}{74.75} \times 35.5=12
\end{aligned}
$$
Valency of metal
$$
\begin{array}{r}
=\frac{2 \times \text { V.D. }}{\text { Equivalent wt. of metal }+35.5} \\
=\frac{2 \times 94.8}{12+35.5}=4
\end{array}
$$
$\therefore$ Formula of compound $=\mathrm{MCl}_4$
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