MCQ
A compound contains $69.5\%$ oxygen and $30.5\%$ nitrogen and its molecular weight is $92$ . The formula of that compound is
- A$N_2O$
- B$NO_2$
- ✓$N_2O_4$
- D$N_2O_5$
$=4.34$
moles of nitrogen $=30.5 / 14$
$=2.18$
ratio of moles of nitrogen and oxygen $=2.18: 4.34$
$=1: 2$
empirical formula $ = NO_2$
molecular formula $=(\mathrm{NO}_ 2) \mathrm{n}$
where $\mathrm{n}=$ molecular mass/ empirical mass
$=92 / 46$
$=2$
therefore the formula of the compound is $\mathrm{N}_ 2 \mathrm{O} _4 .$
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{$E\, =$ Central atom, $B\, =$ Terminal atom, $L\, =$ Lone pair}