In Dumas method for estimation of nitrogen, 0 | vedclass

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(B) Step $1$: Calculate the pressure of dry $N_2$ gas. $P_{N_2} = P_{total} - P_{aqueous} = 715 \ mm - 15 \ mm = 700 \ mm = \frac{700}{760} \ atm$. St

Vedclass · Accepted Answer

$15$

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(B) Step $1$: Calculate the pressure of dry $N_2$ gas. $P_{N_2} = P_{total} - P_{aqueous} = 715 \ mm - 15 \ mm = 700 \ mm = \frac{700}{760} \ atm$. Step $2$: Use the ideal gas equation $PV = nRT$ to find the moles of $N_2$. $n_{N_2} = \frac{PV}{RT} = \frac{(700/760 \ atm) 	imes (0.070 \ L)}{0.0821 \ L \cdot atm \cdot K^{-1} \cdot mol^{-1} 	imes 300 \ K} \approx 0.00263 \ mol$. Step $3$: Calculate the mass of $N_2$. $W_{N_2} = n_{N_2} 	imes 28 \ g/mol = 0.00263 	imes 28 \approx 0.07364 \ g$. Step $4$: Calculate the percentage of nitrogen. $\% \ N = \frac{W_{N_2}}{W_{compound}} 	imes 100 = \frac{0.07364 \ g}{0.50 \ g} 	imes 100 \approx 14.73 \%$. Rounding to the nearest integer, the answer is $15 \% $.

In Dumas method for estimation of nitrogen, $0.50 \ g$ of an organic compound gave $70 \ mL$ of nitrogen collected at $300 \ K$ and $715 \ mm$ pressure. The percentage of nitrogen in the organic compound is . . . . . . $ \% $ (Aqueous tension at $300 \ K$ is $15 \ mm$).

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