Question

a flask contains a mixture of two gases: no2 and ne. the flask holds 0.583 mol of no2 and an unknown amount of ne. if the partial pressure of ne is 0.23 bar and the total pressure in the flask is 1.15 bar, calculate the mass of ne in the flask (in g). tip: you dont need the volume of the flask nor the temperature to solve this problem. answer:

Explanation:

Step1: Find mole - fraction of Ne

The mole - fraction ($x_{Ne}$) of a gas is given by the ratio of its partial pressure ($P_{Ne}$) to the total pressure ($P_{total}$).
$x_{Ne}=\frac{P_{Ne}}{P_{total}}$
Given $P_{Ne} = 0.23$ bar and $P_{total}=1.15$ bar.
$x_{Ne}=\frac{0.23}{1.15}= 0.2$

Step2: Find moles of Ne

Let the moles of Ne be $n_{Ne}$ and moles of $NO_2$ be $n_{NO_2}=0.583$ mol.
The mole - fraction formula is also $x_{Ne}=\frac{n_{Ne}}{n_{Ne}+n_{NO_2}}$.
Since $x_{Ne} = 0.2$, we have $0.2=\frac{n_{Ne}}{n_{Ne}+0.583}$.
$0.2(n_{Ne}+0.583)=n_{Ne}$
$0.2n_{Ne}+0.1166=n_{Ne}$
$0.1166=n_{Ne}-0.2n_{Ne}$
$0.8n_{Ne}=0.1166$
$n_{Ne}=\frac{0.1166}{0.8}=0.14575$ mol

Step3: Calculate mass of Ne

The molar mass of Ne ($M_{Ne}$) is approximately $20.18$ g/mol.
The mass ($m_{Ne}$) of Ne is given by $m_{Ne}=n_{Ne}\times M_{Ne}$
$m_{Ne}=0.14575\times20.18 = 2.94$ g

Answer:

$2.94$ g