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Q.

0.15 gm of an organic compound on combustion gave 0.44 gm of $C O_{2}$ and 0.27gm of $H_{2} O$ .The percentage carbon and hydrogen in the organic compound respectively

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a

90,10

b

60,40

c

80,20

d

70,30

answer is B.

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Detailed Solution

$% o f C = \frac{12}{44} \times \frac{0.44}{0.15} \times 100$

$= 80$

$% o f H = \frac{2}{18} \times \frac{0.27}{0.15} \times 100$

= 20

Question Image

E.F. = $C H_{3}$

$M . F = C_{2} H_{6}$

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