(i) Here, the mass of the substance taken = 0.15g
Mass of AgBr formed = 0.12g
Now 1 mole of AgBr = 1 mole of Br
or (108 + 80) = 188 g of AgBr = 80g of Br
\(\because\) 188 of AgBr contain bromine = 80g
\(\therefore\) 0.12 of AgBr will contain bromine
= \(\frac{80}{188}\) x 0.12g
But this much amount of bromine is present in 0.15g of the organic compound.
Percentage of bromine = \(\frac{80\times0.12\times100}{188\times0.15}\)
= 34.04%
(ii) No, oxygen cannot be detected by Lassaigne’s test.
(iii) Blood red colour.