Solutions Question 213
Question: The normality of 10% (weight/volume) acetic acid is [CPMT 1983]
Options:
A) 1 N
B) 10 N
C) 1.7 N
D) 0.83 N
Show Answer
Answer:
Correct Answer: C
Solution:
$ N=\frac{w\times 1000}{Eq.wt.\times Volume}=\frac{10\times 1000}{60\times 100}=1.66,N $ .