A closed container contains a mixture of non- | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) According to Dalton's Law of Partial Pressures,the total pressure $P_{Total}$ is the sum of partial pressures of individual gases: $P_{Total} = P_

Vedclass · Accepted Answer

$0.45$ and $0.55$

Vedclass · Answer

(C) According to Dalton's Law of Partial Pressures,the total pressure $P_{Total}$ is the sum of partial pressures of individual gases: $P_{Total} = P_A + P_B = 4.5 \ bar + 5.5 \ bar = 10.0 \ bar$. The mole fraction $x_i$ of a gas is given by the ratio of its partial pressure to the total pressure: $x_A = \frac{P_A}{P_{Total}} = \frac{4.5 \ bar}{10.0 \ bar} = 0.45$. Similarly,for gas $B$: $x_B = \frac{P_B}{P_{Total}} = \frac{5.5 \ bar}{10.0 \ bar} = 0.55$. Therefore,the mole fractions of $A$ and $B$ are $0.45$ and $0.55$ respectively.

$A$ closed container contains a mixture of non-reacting gases $A$ and $B$. The partial pressures of $A$ and $B$ are $4.5 \ bar$ and $5.5 \ bar$ respectively. Find the mole fractions of $A$ and $B$ respectively.

Similar Questions

What is the $SI$ unit of rate of diffusion for a gas?

According to Graham's law of diffusion at a given temperature,the ratio of the rate of diffusion $r_A/r_B$ for gases $A$ and $B$ is given by:

The ratio of the rate of diffusion of $SO_2$ to $CH_4$ is

$A$ sample of $N_2$ gas at $0.8 \, atm$ pressure takes $38 \, s$ to effuse through a pinhole. $A$ sample of $1 \, mol$ of an unknown gas '$X$' at $1.6 \, atm$ pressure takes $57 \, s$ to effuse through the same pinhole. The molar mass of gas '$X$' in $g/mol$ is .............

$A$ gaseous mixture contains $65\%$ $N_2$,$15\%$ $O_2$,and $20\%$ $CO_2$ by volume at a total pressure of $760 \, mm \, Hg$. Calculate the partial pressure of each gas in $mm \, Hg$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module