In Dumas method for estimation of nitrogen $0.30 g$ of an organic compound gave $50 m l$ of nitrogen collected at $300 K$ and $715 m m$ of $H g$ pressure. What is the percentage composition of nitrogen in the compound if aqueous tension at $300 K$ is $15 m m$ of $H g$ ?

15.23

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17.46

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19.23

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Solution

The correct option is B $17.46$ %
Volume of Nitrogen collected at $300 K$ and $715 m m$
pressure is $50 m l$
Actual pressure $= 715 - 15 = 700 m m$
Volume of Nitrogen at $S T P = \frac{273 \times 700 \times 50}{300 \times 760} [\frac{P r e s s u r e \times V o l . \times 273}{760 \times T e m p e r a t u r e}]$
$= 41.9 m l$
$22, 400 m l$ of $N_{2}$ at $S T P$ weighs $= 28 g$
$41.9 m l$ of $N_{2}$ weighs $= \frac{28 \times 41.9 g}{22400}$
Percentage of $N_{2} = \frac{28 \times 41.9 \times 100}{22400 \times 0.3} = 17.46$ %

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In Dumas method for estimation of nitrogen 0.30 g of an organic compound gave 50 ml of nitrogen collected at 300 K and 715 mm of Hg pressure. What is the percentage composition of nitrogen in the compound if aqueous tension at 300 K is 15 mm of Hg?

In Dumas method for estimation of nitrogen $0.30 g$ of an organic compound gave $50 m l$ of nitrogen collected at $300 K$ and $715 m m$ of $H g$ pressure. What is the percentage composition of nitrogen in the compound if aqueous tension at $300 K$ is $15 m m$ of $H g$ ?