Question

The $100ml$ of ${H_2}S{O_4}$ solution having molarity $1M$ and density $1.5g{\left( {ml} \right)^{ - 1}}$ is mixed with $400ml$ of water. Calculate final molarity of ${H_2}S{O_4}$ solution, if final density is $1.25g{\left( {ml} \right)^{ - 1}}$ .
(A) $4.4M$
(B) $0.145M$
(C) $0.52M$
(D) $0.227M$

Answer 1

Hint : Density is defined as the mass per unit volume. The molarity of one solution will change when solvents like water are added to the solution. The final molarity can be calculated by taking the total number of moles and total volume of final solution.
$M = \dfrac{n}{V}$
$M$ is molarity
$n$ is number of moles of solution
$V$ is the volume of solution in litres.

Complete Step By Step Answer:
The molarity is a unit used to express the concentration and defined as the number of moles of solute in a given volume of solution in liters. Molarity is also known as molar concentration.
Given that $100ml$ of ${H_2}S{O_4}$ solution having molarity $1M$ and density $1.5g{\left( {ml} \right)^{ - 1}}$ is mixed with $400ml$ of water.
The number of moles of ${H_2}S{O_4}$ is $1M \times 100ml = 1M \times 0.1L = 0.1moles$
The volume of ${H_2}S{O_4}$ solution before addition of water is $100 \times 1.5 = 150ml$ as density is mass per unit volume.
After the addition of water, the total volume will be $150 + 400 = 550ml$
But this final volume has the density of $1.25g{\left( {ml} \right)^{ - 1}}$ . Thus, the final volume will be $\dfrac{{550}}{{1.25}} = 440ml$
Now, the number of moles of ${H_2}S{O_4}$ is $0.1moles$ and the final volume ${H_2}S{O_4}$ solution is $440ml$
The molarity will be $\dfrac{{0.1}}{{400ml}} \times 1000 = 0.227M$
The final molarity of ${H_2}S{O_4}$ solution is $0.227M$
Option D is the correct one.

Note :
While calculating the molarity, the volume must be in litres. If the volume is in milliliters then the molarity should multiply with $1000$ . The density of liquids must be taken into consideration while calculating the molarity. As the density also changes when the solvents are added to the solution.