Use app×
QUIZARD
QUIZARD
JEE MAIN 2026 Crash Course
NEET 2026 Crash Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
127 views
in Chemistry by (81.3k points)
closed by
`3.7 gm` of gas at `25^(@)C` occupied the same volume as `0.184 gm` of hydrogen at `17^(@)C ` and at the same pressure. What is the molecular mass of the gas ?

1 Answer

0 votes
by (75.0k points)
selected by
 
Best answer
For hydrogen,
`w = 0.184 g , T = 17 + 273 = 290 K, M = 2`
We know that, `PV = (w)/(M) RT`
`= (0.184)/(2) xx R xx 290`.....(i)
For unknown gas,
`w = 3.7 g, T = 25 + 273 = 298 K, M = ?`
`PV = (3.7)/(M) xx R xx 298`....(ii)
Equating both the equations,
`(3.7)/(M) xx R xx 298 = (0.184)/(2) xx R xx 290`
or `M = (3.7 xx 298 xx 2)/(0.184 xx 290) = 41.33`

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...