Given below are two statements: one is labelled as Assertion $A$ and the other is labelled as Reason $R$.
Assertion $A$: Helium is used to dilute oxygen in diving apparatus.
Reason $R$: Helium has high solubility in $O_2$.
In the light of the above statements,choose the correct answer from the options given below:

  • A
    $A$ is false but $R$ is true.
  • B
    Both $A$ and $R$ are true and $R$ is the correct explanation of $A$.
  • C
    Both $A$ and $R$ are true and $R$ is $NOT$ the correct explanation of $A$.
  • D
    $A$ is true but $R$ is false.

Explore More

Similar Questions

The values of the Henry's law constant of $Ar$,$CO_{2}$,$CH_{4}$ and $O_{2}$ in water at $25^{\circ} C$ are $40.30$,$1.67$,$0.41$ and $34.86 \ kbar$,respectively. The order of their solubility in water at the same temperature and pressure is

The solubility of a gas in water depends on

The Henry's law constant $(K_H)$ values of three gases $(A, B, C)$ in water are $145$,$2 \times 10^{-5}$,and $35$ kbar,respectively. The solubility of these gases in water follow the order:

Henry's law constants for $O_2$ and $N_2$ are $K_{O_2} = 3.3 \times 10^7 \ atm$ and $K_{N_2} = 6.51 \times 10^7 \ atm$. Calculate the ratio of $\frac{X_{O_2}}{X_{N_2}}$,i.e.,the ratio of mole fractions of $O_2$ and $N_2$ dissolved in water at $25 \ ^\circ C$ from air. (Assume air contains $20 \%$ $O_2$ and $80 \%$ $N_2$ by volume).

$1 \ g$ of silver gets distributed between $10 \ cm^{3}$ of molten zinc and $100 \ cm^{3}$ of molten lead at $800^{\circ} C$. The percentage of silver still left in the lead layer is approximately

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