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# $25g$ of $Mcl_4$ contains $0.5\;mol$ chlorine then its molecular mass is

$\begin{array}{1 1} 100\;g\;mol^{-1} \\ 200\;g\;mol^{-1} \\ 150\;g\;mol^{-1} \\ 400\;g\;mol^{-1}\end{array}$

1 mol of $Mcl_4$ contains 4mol of chlorine.
$\therefore$ 0.5 mol chlorine is present in 25g of $Mcl_4$
$\therefore$ 4 mol chlorine will be present in $\large\frac{25}{0.5}$$\times 4$
(i.e) $200g$ of $Mcl_4$
Hence (b) is the correct option.