AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

Pressure in a bulb dropped from 2000 to 1500 mm in 50 minute when the contained oxygen leaked through a small hole. The bulb was then completely evacuated. A mixture of oxygen and another gas of molecular weight 72 in molar ratio 1:1 at a total pressure of 600 mm was introduced. Find the molar ratio of two gases remaining in the bulb after a period of 70 minute. If your answer is x then fill your OMR with 46 x. (Exclude decimal Place)

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

answer is 39.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

In $(\frac{2000}{1500}) = \frac{k \times 50}{\sqrt{32}}$

In $(\frac{3000}{x}) = k \times \frac{70}{\sqrt{32}}$

$\Rightarrow x = 2000$

$\begin{array}{l} \frac{3000 - 2000}{3000 - P} = \sqrt{\frac{72}{32}} \\ P = 2333.33 \end{array}$

Molar ratio $= \frac{2000}{2333.33} = x$

$46 x ≃ 39$

Watch 3-min video & get full concept clarity

courses

No courses found

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Online Course for IIT JEE

Best Online Course for NEET

Best Online Course for CBSE

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Get Expert Academic Guidance – Connect with a Counselor Today!