question_answer 21)

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

Answer:

Actual pressure of the gas =715 15= 700 mm To convert the volume at experimental conditions to volume at STP AT STP \[~{{P}_{1}}=700\text{ }mm;\] \[~{{P}_{2}}=760\text{ }mm\] \[{{V}_{1}}=50mL;\] \[{{V}_{2}}=?\] \[{{T}_{1}}=300K;\] \[{{T}_{2}}=273K\] Applying general gas equation, \[\frac{{{P}_{1}}{{V}_{1}}}{{{T}_{1}}}=\frac{{{P}_{2}}{{V}_{2}}}{{{T}_{2}}}\] Volume of nitrogen at STP, \[{{V}_{2}}=\frac{{{P}_{1}}{{V}_{1}}}{{{T}_{1}}}\times \frac{{{T}_{2}}}{{{P}_{2}}}\] \[=\frac{700\times 50\times 273}{300\times 760}=41.9mL.\] Weight of organic compound = 0.30 g Percentage of nitrogen in the compound \[=\frac{28}{22400}\times \frac{{{V}_{2}}}{W}\times 100\] \[=\frac{28\times 41.9\times 100}{22400\times 0.30}=\mathbf{17}\mathbf{.46%}\]