Question

In Duma's method, 0.52 g of an organic compound on combustion gave 68.6 mL N₂ at 27°C and 756 mm pressure. What is the percentage of nitrogen in the compound?

Options

• 12.22%

• 14.93%

• 15.84%

• 16.23%

Answer 1

14.93%

Explanation:

V₁ = 68.6 mL, P₁ = 756 mm, T₁ = 300 K

V₂ = ?, P₂ = 760 mm, T₂ = 273 K

`("P"_1"V"_1)/"T"_1 = ("P"_2"V"_2)/"T"_2`

At NTP, vol. of N₂, V₂ = `("P"_1"V"_1)/"T"_1 . "T"_2/"P"_2`

= `(756 xx 68.6)/300 xx 273/760`

= 62.09 mL

Percentage of nitrogen in organic compound = `28/22400 xx "V"_2/"w" xx 100`

= `28/22400 xx 62.09/0.52 xx 100`

= 14.93%