JEE MAIN - Chemistry (2021 - 16th March Evening Shift - No. 20)
In Duma's method of estimation of nitrogen, 0.1840 g of an organic compound gave 30 mL of nitrogen collected at 287 K and 758 mm of Hg pressure. The percentage composition of nitrogen in the compound is __________. (Round off to the Nearest Integer). [Given : Aqueous tension at 287 K = 14 mm of Hg]
Answer
19
Explanation
Aqueous tension at 287 K = 14 mm of Hg.
Hence actual pressure = (758 – 14)
= 744 mm of Hg.
Moles of $${N_2} = {{\left( {758 - 14} \right)} \over {760}} \times {{30 \times {{10}^{ - 3}}} \over {0.0821 \times 287}}$$
$$ = 1.246 \times {10^{ - 3}}$$ mol
mass of $${N_2} = 1.246 \times {10^{ - 3}} \times 28$$
mass % of N2 $$ = {{mass\,of\,'N'} \over {total\,mass}} \times 100$$
$$ = {{1.246 \times 28 \times {{10}^{ - 3}}} \over {0.184}} \times 100$$
$$ = {{124.6 \times 28} \over {0.184}}\% = 18.96\% $$
$$ \simeq 19\% $$
Hence actual pressure = (758 – 14)
= 744 mm of Hg.
Moles of $${N_2} = {{\left( {758 - 14} \right)} \over {760}} \times {{30 \times {{10}^{ - 3}}} \over {0.0821 \times 287}}$$
$$ = 1.246 \times {10^{ - 3}}$$ mol
mass of $${N_2} = 1.246 \times {10^{ - 3}} \times 28$$
mass % of N2 $$ = {{mass\,of\,'N'} \over {total\,mass}} \times 100$$
$$ = {{1.246 \times 28 \times {{10}^{ - 3}}} \over {0.184}} \times 100$$
$$ = {{124.6 \times 28} \over {0.184}}\% = 18.96\% $$
$$ \simeq 19\% $$
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