Question

In a Duma's nitrogen estimation 0.3 g of an organic compound gave 50 mL of nitrogen collected at 300 K and 715 mm pressure. Calculate the percentage of nitrogen in the compound.

Options

• 23.7%

• 17.9%

• 27.4%

• 12.7%

Answer 1

17.9%
Explanation:

To calculate the volume of N₂ at S.Т.Р.
V₁ = 50 mL V₂ = ?
P₁ = 715 mm P₂ = 760 mm
T₁ = 300 K
T₂ = 273 K
Applying \[\frac{p_1V_1}{T_1}=\frac{p_2V_2}{T_2}\]
or   \[V_2=\frac{p_1V_1T_2}{p_2T_1}\]
\[V_2=\frac{715\times50\times273}{760\times300}=42.80\mathrm{mL}\]
22400 mL of nitrogen at S.T.P. weight = 28 g
41.9mLofnitrogenS.T.P.weight = \[\frac{28\times42.80}{22400}=0.0535\mathrm{~g}\]
Percentage of nitrogen = \[\frac{0.0535}{0.3}\times100=17.83\%\]