MCQ
The molality of $90\%$ ${H_2}S{O_4}$ solution is [density $=1.8\,gm/ml$ ]
- A$1.8$
- B$48.4$
- ✓$9.18$
- D$94.6$
Weight of one litre of solution $ = 1800\,gm$
$\therefore $ Weight of ${H_2}S{O_4}$ in the solution$ = \frac{{1800 \times 90}}{{100}}$$=162\,gm$
$\therefore $ Weight of solvent $ = 1800 - 1620 = 180\,gm$
$\therefore $ Molality $ = \frac{{1620}}{{98}} \times \frac{{100}}{{180}} = 9.18$
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$A, C$ and $D$ are :-
