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Question: A bottle contains 100 cc of 0.1m HCl solution. If 900 cc of distilled water is added it, then the co...

A bottle contains 100 cc of 0.1m HCl solution. If 900 cc of distilled water is added it, then the concentration of 100ml of final solution will be:
a. - 0.011 M
b. - 0.11 M
c. - 0.01 M
d. - 0.90 M

Explanation

Solution

Use the formula of dilution i.e. M1V1=M2V2{{M}_{1}}{{V}_{1}}\,\,=\,\,{{M}_{2}}{{V}_{2}} to obtain the concentration of final solution prepared.

Complete Solution :
When we add water to a solution of certain concentration then this process is called dilution. Formula used to calculate missing identity during the process of dilution is M1V1=M2V2{{M}_{1}}{{V}_{1}}\,\,=\,\,{{M}_{2}}{{V}_{2}}.
Here M1{{M}_{1}} = Molarity of given solution (M),
V1{{V}_{1}} = volume of given solution taken (cc),
M2{{M}_{2}} = molarity of solution to be prepare (M),
V2{{V}_{2}} = volume of solution to be prepared (cc).

In given question the values are:
M1{{M}_{1}} = Concentration to HCl taken = 0.1M
V1{{V}_{1}} = volume of HCl taken = 100cc
M2{{M}_{2}} = Concentration of final prepared HCl = ?
V2{{V}_{2}} = Final volume of solution after addition or water = 100cc + 900cc
= 1000 cc

Now, the new concentration can be calculated:
M1V1=M2V2{{M}_{1}}{{V}_{1}}\,\,=\,\,{{M}_{2}}{{V}_{2}}
Rewrite the formula-
M2=M1V1V2{{M}_{2}} = \dfrac{{{M}_{1}}{{V}_{1}}}{{{V}_{2}}}
Put the values in formula
M2{{M}_{2}} = 0.1×1001000\dfrac{0.1\times 100}{1000}​= 0.01M
Now, if 100ml of final solution is taken, the morality of it will be same as the whole solution prepared i.e. 0.01 M.
So, the correct answer is “Option C”.

Note: Firstly, focus on the final solution volume. Another point to be kept in mind is that concentration does not change whatever quantity of solution we take from the final solution.