Equal masses of methane (CH_4) and oxygen (O_ | vedclass

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

Question

(C) Let the mass of both $CH_4$ and $O_2$ be $w \, g$. Number of moles of $CH_4$ $(n_{CH_4})$ = $\frac{w}{16}$. Number of moles of $O_2$ $(n_{O_2})$ =

Vedclass · Accepted Answer

$1/3$

Vedclass · Answer

(C) Let the mass of both $CH_4$ and $O_2$ be $w \, g$. Number of moles of $CH_4$ $(n_{CH_4})$ = $\frac{w}{16}$. Number of moles of $O_2$ $(n_{O_2})$ = $\frac{w}{32}$. According to Dalton's Law of partial pressures,the partial pressure of a gas is equal to its mole fraction multiplied by the total pressure. Mole fraction of $O_2$ $(x_{O_2})$ = $\frac{n_{O_2}}{n_{CH_4} + n_{O_2}} = \frac{w/32}{w/16 + w/32}$. $x_{O_2} = \frac{w/32}{2w/32 + w/32} = \frac{w/32}{3w/32} = \frac{1}{3}$. Therefore,the fraction of total pressure exerted by oxygen is $1/3$.

Equal masses of methane $(CH_4)$ and oxygen $(O_2)$ are mixed in an empty container at $25\,^oC$. The fraction of the total pressure exerted by oxygen is:

Similar Questions

How many grams of $60\% \, H_2SO_4$ are required for the complete reaction of $100 \, g \, CaCO_3$?

$1 \ g$ of a carbonate $(M_2CO_3)$ on treatment with excess $HCl$ produces $0.01 \ mol$ of $CO_2$. The molar mass of $M_2CO_3$ is $.......... \ g \ mol^{-1}$. (Nearest integer)

The number of moles of $Br_2$ produced when two moles of potassium permanganate are treated with excess potassium bromide in aqueous acid medium is

What is the volume occupied by $2 \ g$ of helium gas at $STP$ (in $dm^3$)? (molar mass of helium gas $= 4 \ g \ mol^{-1}$)

Concentrated aqueous sulfuric acid is $98\%$ $H_2SO_4$ by weight and has a density of $1.80 \ g \ mL^{-1}$. The volume of acid required to make $1 \ L$ of $0.1 \ M$ $H_2SO_4$ solution is ....... $mL$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module