MCQ
$120\,gm$ of urea are present in $5\,litre$ solution, the active mass of urea is
- A$0.2$
- B$0.06$
- ✓$0.4$
- D$0.08$
✓
Answer
Correct option: C.
$0.4$
(c) Active mass$ = \frac{{moles}}{{litre}}$
$=\frac{{wt.in\,\, gm / molecular \,\,wt}}{{V\,\, in \,\,liter}}$$=\frac{{120/60}}{{5}}$$=\frac{{2}}{{5}}$$=4$
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