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The amount of bromine required to completely convert of phenol into 2,4,6 tribromophenol is…
The amount of bromine required to completely convert $9\cdot4\text{ g}$ of phenol into 2,4,6 tribromophenol is:
(Atomic weight of $\text{C} = 12$, $\text{H} = 1$, $\text{O} = 16$ and $\text{Br} = 80$)
- (a)$24\text{ g}$
- (b)$48\text{ g}$
- (c)$80\text{ g}$
- (d)$96\text{ g}$
Answer
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From ISC 2025 Improvement Chemistry Paper 1, question 1(B)(iii).