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 $ .



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