In Dumas' method for the estimation of nitrogen, of an organic compound gave of nitrogen gas collect — Qualitative and Quantitative Analysis Chemistry Question
Question
In Dumas' method for the estimation of nitrogen, $0.25 \text{ g}$ of an organic compound gave $44 \text{ mL}$ of nitrogen gas collected over water at $300 \text{ K}$ and $725 \text{ mm Hg}$ pressure. If the aqueous tension at $300 \text{ K}$ is $25 \text{ mm Hg}$, what is the percentage of nitrogen in the compound?
Answer: B
💡 Solution & Explanation
The actual pressure of dry $N_2$ is $725 - 25 = 700 \text{ mm Hg}$. Using the gas law to find volume at STP: $V_{STP} = \frac{273 \times 700 \times 44}{300 \times 760} = 33.52 \text{ mL}$. The weight of $N_2$ is $\frac{28 \times 33.52}{22400} = 0.0419 \text{ g}$. Percentage of $N = (\frac{0.0419}{0.25}) \times 100 = 16.76\%$.
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