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16g $SO_x$ occupies 5.6 litres at STP. Assuming the nature of the gas to be ideal, the value of $x$ is

$\begin{array}{1 1}(a)\;1\qquad(b)\;2\qquad(c)\;3\qquad(d)\;None\;of\;these\end{array}$

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5.6L by volume of $SO_x=16g$ of $SO_x$
$\therefore$ 22.4L $SO_x$ by volume $=\large\frac{16\times 22.4}{5.6} = 64$
Molar weight of $SO_x$ gas = $64$ g
From the chemical formula of $SO_x$, we obtain, $32+x\times 16=64$
$\therefore x=2$
Hence (b) is the correct option.
answered Oct 25, 2013 by sreemathi.v
edited Mar 18, 2014 by mosymeow_1

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