Question

In Dumas method for the estimation of nitrogen, 0.25 g of an organic compound gave 40 mL of nitrogen collected at 300 K temperature and 725 mm of Hg pressure. If the aqueous tension at 300 K is 25 mm of Hg, what is the percentage of nitrogen in the compound (approximately)?

• A. 15.76
• B. 17.36
• C. 18.2
• D. 16.8

Answer 1

Correct option: (C)

Step 1: Calculating the moles of nitrogen
Pressure of pure $N_2$ (excluding pressure from the water vapour)$=725-25=700mmofHg$
Using the ideal gas equation,
$PV=nRT$

$\text{P=Pressure of ideal gas}$

$\text{V=Volume of ideal gas(L)}$

$\text{n=number of moles of ideal gas}$

$\text{R=Gas constant}$

$\text{T=Temperature of gas(K)}$

$\frac{700}{760}atm\times 40\times {10}^{-3}L$= $n\times 0.0821atmLmo{l}^{-1}{K}^{-1}\times 300K$
$n=1.5\times {10}^{-3}mol$

Step 2: Calculating the percentage of nitrogen in given compound

Molecular mass of nitrogen $=2\left(14\right)=28g$

$\text{Moles}=\frac{\text{Given mass}}{\text{Molecular mass}}$

mass of $N_2$ is= $1.5\times 28\times {10}^{-3}=0.042g$

So, the percentage of Nitrogen = $\frac{0.042}{0.25}\times 100=16.8$%

Hence, the correct option is (C).