To 20 mL of 1.5 M $H C l$ solution, 30 mL of water is added and to this resultant solution again 75 mL of 0.4 M HCl is added. The molarity of the resultant solution is:

0.40 M

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2.10 M

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0.56 M

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5.5 M

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Solution

The correct option is C 0.56 M
In the first case,
$M_{1} V_{1} = M_{2} V_{2}$
here, $M_{1} = 1.5 M; V_{1} = 20 m L; V_{2} = 50 m L$
So $M_{2} = \frac{1.5 \times 20}{50} = 0.6 M$
Now, to this 0.6 M of 50 mL $H C l$ solution, 75 mL of 0.4 M $H C l$ is added.
Using,
$M_{3} = \frac{M_{1} V_{1} + M_{2} V_{2}}{V_{3}}$
where $V_{3} = 50 + 75 = 125$ mL
$M_{3} = \frac{(0.6 \times 50) + (75 \times 0.4)}{125} M_{3} = 0.56 M$

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