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In Dumas' method of estimation of nitrogen 0.35 g of an organic compound gave 55 mL of nitrogen collected at 300 K temperature and 715 mm pressure. The percentage composition of nitrogen in the compound would be (Aqueous tension at 300 K = 15 mm)
A
14.45
B
15.45
C
16.45
D
17.45
Detailed Solution
Pressure of dry nitrogen = 715 − 15 = 700 mm Hg
Convert the volume to STP using $\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$
$V_2 = \frac{P_1V_1T_2}{T_1P_2} = \frac{700\times55\times273}{300\times760}$
$V_2 = 46.09$ mL
22400 mL of $N_2$ at STP weighs 28 g, so mass of nitrogen = $\frac{28\times46.09}{22400}$ g = 0.0576 g
% of N = $\frac{28\times46.09\times100}{22400\times0.35}$
% of N = 16.45
Convert the volume to STP using $\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$
$V_2 = \frac{P_1V_1T_2}{T_1P_2} = \frac{700\times55\times273}{300\times760}$
$V_2 = 46.09$ mL
22400 mL of $N_2$ at STP weighs 28 g, so mass of nitrogen = $\frac{28\times46.09}{22400}$ g = 0.0576 g
% of N = $\frac{28\times46.09\times100}{22400\times0.35}$
% of N = 16.45