Some Basic Concepts of Chemistry - Result Question 13
16. 3 g of activated charcoal was added to 50 mL of acetic acid solution $(0.06 \mathrm{~N})$ in a flask. After an hour it was filtered and the strength of the filtrate was found to be 0.042 N . The amount of acetic acid adsorbed (per gram of charcoal) is
(2015 JEE Main)
(a) 18 mg
(b) 36 mg
(c) 42 mg
(d) 54 mg
Show Answer
Answer:
Correct Answer: 16. (d)
Solution:
- Given, initial strength of acetic acid $=0.06 \mathrm{~N}$ Final strength $=0.042 \mathrm{~N} ; \quad$ Volume $=50 \mathrm{~mL}$
$\therefore$ Initial millimoles of $\mathrm{CH}_3 \mathrm{COOH}=0.06 \times 50=3$
Final millimoles of $\mathrm{CH}_3 \mathrm{COOH}=0.042 \times 50=2.1$
$\therefore$ Millimoles of $\mathrm{CH}_3 \mathrm{COOH}$ adsorbed $=3-2.1=0.9 \mathrm{mmol}$
$=0.9 \times 60 \mathrm{mg}=54 \mathrm{mg}$
BETA
