Respuesta :
Answer : The mass of nitrogen present are, 4.965 grams.
Explanation :
According to the Raoult's law,
[tex]p_{He}=X_{He}\times p_T[/tex]
where,
[tex]p_{He}[/tex] = partial pressure of gas = 231 mmHg
[tex]p_T[/tex] = total pressure of gas = 641 mmHg
[tex]X_{He}[/tex] = mole fraction of helium gas = ?
Now put all the given values in this formula, we get the mole fraction of helium gas.
[tex]231mmHg=X_{He}\times 641mmHg[/tex]
[tex]X_{He}=0.36[/tex]
Now we have to calculate the mole fraction of nitrogen gas.
[tex]X_{He}+X_{N_2}=1[/tex]
[tex]X_{N_2}=1=0.36=0.64[/tex]
Now we have to calculate the mass nitrogen gas.
[tex]\frac{X_{He}}{X_{N_2}}=\frac{n_{He}}{n_{N_2}}[/tex]
[tex]\frac{X_{He}}{X_{N_2}}=\frac{\frac{w_{He}}{M_{He}}}{\frac{w_{N_2}}{M_{N_2}}}[/tex]
where,
n = moles, w = mass, M = molar mass
Now put all the given values in this expression, we get:
[tex]\frac{0.36}{0.64}=\frac{\frac{0.399}{4}}{\frac{w_{N_2}}{28}}[/tex]
[tex]w_{N_2}=4.965g[/tex]
Therefore, the mass of nitrogen present are, 4.965 grams.
