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In Duma's method 0.52g of an organic com...

In Duma's method 0.52g of an organic compound on combustion gave 68.6 mL `N_(2)` at `27^(@)C and 756mm` pressure. What is the percentage of nitrogen in the compound?

A

0.1222

B

0.1493

C

0.1584

D

0.1623

Text Solution

Verified by Experts

The correct Answer is:
B
`V_(1)=68.6mL,P_(1)=756mm,T_(1)=300K`
`V_(2)=?,P_(2)=760mm,T_(2)=273K`
`(P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2))`
At NTP, vol. of `N_(2),V_(2)=(P_(1)V_(1))/(T_(1))*(T_(2))/(P_(2))=(756xx68.6)/(300)xx(273)/(760)`
=62.09 mL
Percentage of nitrogen in organic compound
`=(28)/(22400)xx(V_(2))/(w)xx100=(28)/(22400)xx(62.09)/(0.52)xx100=14.93%`
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