Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Medical
  • 200s was needed to effuse ozonise oxygen containing 60% oxygen by volume. Under similar conditions, same amount of oxygen took only 180s. calculate the value of <img alt="\rho" src="https

200s was needed to effuse ozonise oxygen containing 60% oxygen by volume. Under similar conditions, same amount of oxygen took only 180s. calculate the value of \rho of O_3  if for O_2 , \rho \text { is } 1.6 \mathrm{~g} / \mathrm{L} \text {. }

Option: 1

1.936 \mathrm{~g} / \mathrm{L}


Option: 2

2.16 \mathrm{~g} / \mathrm{L}


Option: 3

2.52 \mathrm{~g} / \mathrm{L}


Option: 4

3.28 \mathrm{~g} / \mathrm{L}


Answers (1)

 Let} \mathrm{V_{M L}} of gas effused 

\mathrm{\begin{aligned} \frac{v / 200}{v / 180} & =\sqrt{\frac{d_{D_2}}{d_{\text {mix }}}} \\ \frac{180}{200} & =\sqrt{\frac{d_{\theta_2}}{d_{\text {mix }}}} \\ d_{\text {mix }} & =1.6 \times\left(\frac{20}{18}\right)^2 \\ & =1.97 \end{aligned}}

Let \mathrm{\mathrm{\rho}_{\mathrm{O}_3}} is d

In 100 volume ozonise oxygen \mathrm{60 \% \mathrm{O}_2} &

\mathrm{40 \% \text { by } \vee \mathrm{O}_3 \text { is present. }}

\mathrm{\therefore \text { Mass of mixture }=\text { mass of } \mathrm{O}_3+\text { mass of } \mathrm{O}_2}

\mathrm{40 \% \text { by } \vee \mathrm{O}_3} is present

\mathrm{\begin{aligned} 100 \times 1.97 & =(40 d)+(60 \times 1.6) \\ 197 & =40 d+96 \\ d & =2.52 . \end{aligned}}

 

Posted by

Ramraj Saini

View full answer

NEET 2024 Most scoring concepts

    Just Study 32% of the NEET syllabus and Score up to 100% marks

Similar Questions

Trending Articles/News

anam-khan

Reminder! NEET UG 2025 registration ends in two days; qualifying exam code
anam-khan

NEET UG 2025 chemistry chapter-wise weightage; high-scoring topics; sectional breakdown
anam-khan

NEET UG registration 2025 will close in a week, NTA advises students to ‘avoid last minute rush’

Latest Question