Calculate the resultant molarity on mixing $2500 m L$ of a $2 M H N O_{3}$ solution and $189 g$ of $H N O_{3}$ .

3.2 M

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4.7 M

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1.2 M

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5 M

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Solution

The correct option is A $3.2 M$
$Molarity = \frac{number of moles of solute}{volume of solution in mL} \times 1000$

$Initial number of moles of solute = \frac{2 \times 2500}{1000} = 5 m o l$

$Moles of H N O_{3} added = \frac{189}{63} = 3 m o l$

$Total moles = 5 + 3 = 8 mol$

$New molarity = \frac{8}{2500} \times 1000 = 3.2 M$

Molarity after addition of solute will be $3.2 M$ .

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