A bottle contains 100cc of 0.1 m HCl solution. If 900 cc of distilled water is added to it, then the concentration of 100 ml of the final solution will be:

0.011 M

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0.11 M

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0.01 M

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0.90 M

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Solution

The correct option is C 0.01 M
Given,
Molarity of $H C l = 0.1 M$
Volume of $H C l = 100 c c = 100 m l$
Distilled water volume= $900 c c = 900 m l$
Now, the new concentration is calculated by
$M_{1} V_{1} = M_{2} V_{2}$
Here $V_{2} = 100 m l + 900 m l$ =Final Volume
$M_{2}$ =Final concentration
$\Rightarrow M_{2} = \frac{M_{1} V_{1}}{V_{2}} = \frac{0.1 \times 100}{1000} = 0.01 M$
Now, if $100 m l$ of final solution is taken, the molarity will remain same
$∴$ Molarity of $100 m l$ of final solution= $0.01 M$

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