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Quantitative analysis
Appears in
Concepts tested here
- Dumas method 2
- Kjeldahl's method
All Questions
2015 AIPMT-I 1 question
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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:$V_1 = 40$ 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 $= \frac{28\times 33.52\times 100}{22400\times 0.25} = 16.76$
2014 AIPMT 1 question
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In the Kjeldahl's method for estimation of nitrogen present in a soil sample, ammonia evolved from 0.75 g of sample neutralized 10 mL of 1 M $H_2SO_4$. The percentage of nitrogen in the soil is:10 mmol $H_2SO_4$ neutralises 20 mmol $NH_3$: $H_2SO_4 + 2NH_3 \rightarrow (NH_4)_2SO_4$
$20\times10^{-3}$ mol $NH_3$ contains $14\times20\times10^{-3}$ g nitrogen.
% nitrogen $= \frac{14\times20\times10^{-3}}{0.75}\times100 = 37.33\%$
2011 AIPMT-PRE 1 question
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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)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