Solutions Question 408

Question: 1.0 gm of pure calcium carbonate was found to require 50 ml of dilute $ HCl $ for complete reaction. The strength of the $ HCl $ solution is given by [CPMT 1986]

Options:

A) 4 N

B) 2 N

C) 0.4 N

D) 0.2 N

Show Answer

Answer:

Correct Answer: C

Solution:

M.eq. of HCl = M.eq. of $ CaCO_3 $

$ N\times 50=\frac{1}{50}\times 1000 $ ; $ N=\frac{1\times 1000}{50\times 50}=0.4N $



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