MCQ
The total number of gm-molecules of $S{O_2}C{l_2}$ in $13.5\,g$ of sulphuryl chloride is
- ✓$0.1$
- B$0.2$
- C$0.3$
- D$0.4$
✓
Answer
Correct option: A.
$0.1$
a
(a) Molecular weight of $S{O_2}C{l_2}$
(a) Molecular weight of $S{O_2}C{l_2}$
$ = 32 + 32 + 2 \times 35.5$= $135\,gm$
$135\, gm$ of $S{O_2}C{l_2}$ $=1\,gm$ molecule
$\therefore $ $13.5\,gm$ of $S{O_2}C{l_2}$ $ = \frac{1}{{135}} \times 13.5 = 0.1$.
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