Air contains $23\% \, O_2$ and $77\% \, N_2$ by weight. What is the percentage of $O_2$ by volume?

  • A
    $28.1$
  • B
    $20.8$
  • C
    $21.8$
  • D
    $23$

Explore More

Similar Questions

The ratio of $C_p$ and $C_v$ of a gas '$X$' is $1.4$. The number of atoms of the gas '$X$' present in $11.2 \ L$ of it at $N.T.P.$ is

The normality of $H_{2}SO_{4}$ in the solution obtained on mixing $100 \ mL$ of $0.1 \ M \ H_{2}SO_{4}$ with $50 \ mL$ of $0.1 \ M \ NaOH$ is $\times 10^{-1} \ N$. (Nearest Integer)

What is $Rasavidya$?

Match the following columns:
Column - $A$ Column - $B$
$(A)$ $88 \ g$ of $CO_2$ $(1)$ $0.2 \ mol$
$(B)$ $6.022 \times 10^{23}$ molecules of water $(2)$ $2 \ mol$
$(C)$ $5.6 \ L$ of $O_2$ at $STP$ $(3)$ $1 \ mol$
$(D)$ $96 \ g$ of $O_2$ $(4)$ $6.022 \times 10^{23}$ molecules
$(E)$ One mole of any gas at $STP$ $(5)$ $3 \ mol$

An athlete is given $100 \ g$ of glucose $(C_6H_{12}O_6)$ for energy. This is equivalent to $1800 \ kJ$ of energy. The $50 \ \%$ of this energy gained is utilized by the athlete for sports activities at the event. In order to avoid storage of energy,the weight of extra water he would need to perspire is $......... \ g$ (Nearest integer).
Assume that there is no other way of consuming stored energy. Given: The enthalpy of evaporation of water is $45 \ kJ \ mol^{-1}$.
Molar mass of $C, H$ and $O$ are $12, 1$ and $16 \ g \ mol^{-1}$.

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo