Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
80 views
in Chemistry by (89.4k points)
closed by
`3g` of actived chacoal was added to 50mL of acetic acid solution `(0.06N)` in a flask. After an hour it was filterred and the strength of the filtrate was found to be `0.042N` . The amount of acetic adsorbed (per gram of charcoal) is:
A. `18 mg`
B. `36 mg`
C. `42 mg`
D. `54 mg`

1 Answer

0 votes
by (88.1k points)
selected by
 
Best answer
Correct Answer - D
Given, initial strength of acetic acid `= 0.06 N`
Final strength `= 0.042 N`, Volume `= 50 mL`
`:.` Initial millimoles of `CH_(3)COOH = 0.06 xx 50 = 3`
Final millimoles of `CH_(3)COOH` adsorbed `= 3-2.1 = 0.9 mmol = 0.9 xx 60 mg = 54 mg`

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

...