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In Dumas' method for estimation of nitrogen, 0.25 g of an organic compound gave 40 mL of nitrogen collected at 300 K temperature and 725 mm pressure. If the aqueous tension at 300 K is 25 mm, the percentage of nitrogen in the compound is:
A
17.36
B
18.20
C
16.76
D
15.76
Detailed Solution
$V_1 = 40$ mL, $T_1 = 300$ K, $P_1 = 725 - 25 = 700$ mm Hg
Volume at STP: $V_2 = \frac{P_1V_1T_2}{T_1P_2} = \frac{700\times 40\times 273}{300\times 760} = 33.52$ mL
% of N $= \frac{28\times V\times 100}{22400\times \text{mass of compound}}$
% of N $= \frac{28\times 33.52\times 100}{22400\times 0.25} = 16.76$
Volume at STP: $V_2 = \frac{P_1V_1T_2}{T_1P_2} = \frac{700\times 40\times 273}{300\times 760} = 33.52$ mL
% of N $= \frac{28\times V\times 100}{22400\times \text{mass of compound}}$
% of N $= \frac{28\times 33.52\times 100}{22400\times 0.25} = 16.76$