(i) In Carius method of estimation of halogen, 0.15 g of an organic co

1.	(i) In Carius method of estimation of halogen, 0.15 g of an organic compound gave 0.12 g of AgBr, find out the % of Br in the compound. (At. mass of Ag = 108, Br = 80)(ii) Can we detect oxygen in a compound by Lassaigne's test?(iii) Which colour will appear in the Lassaigne's test if the compound contains both nitrogen and Sulphur?
Answer» (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.

(i) In Carius method of estimation of halogen, 0.15 g of an organic compound gave 0.12 g of AgBr, find out the % of Br in the compound. (At. mass of Ag = 108, Br = 80)(ii) Can we detect oxygen in a compound by Lassaigne's test?(iii) Which colour will appear in the Lassaigne's test if the compound contains both nitrogen and Sulphur?

Answer»

(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.