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Q.

In the estimation of nitrogen by Dumas method 0.112 gm of an organic compound gave 22.4 ml of nitrogen gas at STP. The percentage of nitrogen in the compound is

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a

15.5

b

20.2

c

47.5

d

25.0

answer is C.

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Detailed Solution

Moles of $N_{2} = \frac{22 .4}{22400} = 10^{- 3}$ Wt $N_{2} = 10^{- 3} \times 28$ $% of N_{2} = \frac{10^{- 3} \times 28}{0 .112} \times 100$ = 25

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