In Dumas' method for estimation of nitrogen, $0.3 g$ of an organic compound gave $50 m L$ of nitrogen collected at $300 K$ temperature and $715 m m$ pressure. Calculate the percentage composition of nitrogen in the compound. (Aqueous tension at $300 K = 15 m m$ )

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Solution

Given,
The volume of nitrogen collected at $300 K$ and $715 m m$ pressure is $50 m L$

But, Actual pressure $= 715 - 15 = 700 m m$

Volume nitrogen at STP $= \frac{273 \times 700 \times 50}{300 \times 760}$

$= 41.9 m L$

We know,
$22, 400 m L$ of nitrogen at STP weighs $= 28 g$

$∴ 41.9 m L$ of nitrogen weighs $= \frac{28 \times 41.9}{22400} g$

percentage of nitrogen $= \frac{28 \times 41.9 \times 100}{22400 \times 0.3} = 17.46 %$

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In Dumas' method for estimation of nitrogen, 0.3g of an organic compound gave 50mL of nitrogen collected at 300K temperature and 715mm pressure. Calculate the percentage composition of nitrogen in the compound. (Aqueous tension at 300K=15mm)

In Dumas' method for estimation of nitrogen, $0.3 g$ of an organic compound gave $50 m L$ of nitrogen collected at $300 K$ temperature and $715 m m$ pressure. Calculate the percentage composition of nitrogen in the compound. (Aqueous tension at $300 K = 15 m m$ )