
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
Answer
537k+ views
Hint: Use the formula of dilution i.e. ${{M}_{1}}{{V}_{1}}\,\,=\,\,{{M}_{2}}{{V}_{2}}$ 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 ${{M}_{1}}{{V}_{1}}\,\,=\,\,{{M}_{2}}{{V}_{2}}$.
Here ${{M}_{1}}$ = Molarity of given solution (M),
${{V}_{1}}$ = volume of given solution taken (cc),
${{M}_{2}}$ = molarity of solution to be prepare (M),
${{V}_{2}}$ = volume of solution to be prepared (cc).
In given question the values are:
${{M}_{1}}$ = Concentration to HCl taken = 0.1M
${{V}_{1}}$ = volume of HCl taken = 100cc
${{M}_{2}}$ = Concentration of final prepared HCl = ?
${{V}_{2}}$ = Final volume of solution after addition or water = 100cc + 900cc
= 1000 cc
Now, the new concentration can be calculated:
${{M}_{1}}{{V}_{1}}\,\,=\,\,{{M}_{2}}{{V}_{2}}$
Rewrite the formula-
${{M}_{2}} = \dfrac{{{M}_{1}}{{V}_{1}}}{{{V}_{2}}}$
Put the values in formula
${{M}_{2}}$ = $\dfrac{0.1\times 100}{1000}$= 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.
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 ${{M}_{1}}{{V}_{1}}\,\,=\,\,{{M}_{2}}{{V}_{2}}$.
Here ${{M}_{1}}$ = Molarity of given solution (M),
${{V}_{1}}$ = volume of given solution taken (cc),
${{M}_{2}}$ = molarity of solution to be prepare (M),
${{V}_{2}}$ = volume of solution to be prepared (cc).
In given question the values are:
${{M}_{1}}$ = Concentration to HCl taken = 0.1M
${{V}_{1}}$ = volume of HCl taken = 100cc
${{M}_{2}}$ = Concentration of final prepared HCl = ?
${{V}_{2}}$ = Final volume of solution after addition or water = 100cc + 900cc
= 1000 cc
Now, the new concentration can be calculated:
${{M}_{1}}{{V}_{1}}\,\,=\,\,{{M}_{2}}{{V}_{2}}$
Rewrite the formula-
${{M}_{2}} = \dfrac{{{M}_{1}}{{V}_{1}}}{{{V}_{2}}}$
Put the values in formula
${{M}_{2}}$ = $\dfrac{0.1\times 100}{1000}$= 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.
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