JEE MainChemistryModeratePhysical Chemistry

A gaseous mixture of three gases A, B and C has a pressure of 10
atm. The total number of moles of all the gases is 10. If the partial
pressure of A and B are 3.0 and 1.0 atm respectively and if C has mol.
wt. of 2.0, what is the weight of C in g present in the mixture ?

Show answer and solution

Pressure of Total mixture = 10 atm

P_{A} + P_{B} + P_{c} = 10

3 + 1 + P_{c} = 10 \Rightarrow P_{c} = 6\ atm\

Total moles of mixture = 10

N_{A} + n_{B} + n_{C} = 10

\frac{P_{A}}{P_{B}} = \frac{n_{A}}{n_{B}} = \frac{3}{1} \Rightarrow \frac{P_{B}}{P_{C}} = \frac{n_{B}}{n_{C}} = \frac{1}{6}

Let

n_{A} = K \Rightarrow n_{B} = \frac{K}{3}n_{c} = \frac{1}{6}n_{B} = 2K

\Rightarrow K + \frac{K}{3} + 2K = 10 \Rightarrow \frac{K}{3} = \frac{n_{C}}{6} \Rightarrow n_{c} = 2K

\Rightarrow K\left( \frac{10}{3} \right) = 10K = 3, \Rightarrow n_{A} = 3

n_{B} = 1

n_{c} = 6

weight of in mixture =.

\ = 2 \times 6 = 12

Source: Grass-G / CH-23 Chemistry Paper 31 Dec. FST.docx

Need a video for this idea?

Scan any doubt from your notes and get a whiteboard explanation.