In Duma's method for estimation of nitrogen `0.25 g` of an organic compound gave `40mL` of nitrogen collected at `300K` temperature of `725 mm` pressure. If the aqueous tension at `300K` is `25 mm`, the percentage of nitrogen in the compound is

Options

A. `16.76`
B. `15.76`
C. `17.36`
D. `18.20`

Answer

A

Explanation

Vapour pressure of gas `= 725 - 25 = 700` mm Hg
To calculate volume at S. T. P.
`{:(V_1= 40 "mL", V_2 = ?),(P_1= 700 "mm",P_2 = 760 "mm"),(T_1 = 300K, T_2 = 273 K):}`
Applying `(P_1 V_1)/( T_1) = (P_2 V_2)/( T_2)`
or `V_1 = (P_1 V_1 T_2)/( P_2 T_1) = ( 700 xx 40 xx 273)/( 760 xx 300)`
`= 33.53` mL
22400 mL of `N_2` at S. T. P. weigh = 28 g
33.53 mL of `N_2` at S. T. P. weigh `= (28) /( 22400) xx 33.53`
`= 0.0419` g
`%` of nitrogen `= ( 0.0419)/( 0.25) xx 100`
`= 16.76 %`