Find the molarity of a solution made by mixing equal volumes of 30% by weight of $H_{2}S{O}_4$ (density = 1.218 g/mL) and 70% by weight of $H_{2}S{O}_4$ (density = 1.610 g/mL) solutions.

The correct option is B 7.61 M
Assuming V mL of $H_{2}S{O}_4$ is mixed together.
For 30% (w/w) solution:
30% (w/w) means 30 g of solute is present in 100 g of the solution
Since, density of solution = $\frac{\text{mass of the solution}}{\text{volume of the solution}}$
Mass of $H_{2}S{O}_4$ solution = 1.218 $\times V$ = 1.218V g
100 g of the solution contains 30 g of $H_{2}S{O}_4$
1.218V g of the solution will contain = $\frac{30}{100}\times 1.218V$ = 0.3654V g...(i)
Similarly,
70% (w/w) means 70 g of solute is present in 100 g of the solution
Since, density of solution = $\frac{\text{mass of the solution}}{\text{volume of the solution}}$
Mass of $H_{2}S{O}_4$ solution = 1.610 $\times V$ = 1.610V g
100 g of the solution contains 70 g of $H_{2}S{O}_4$
1.610V g of the solution will contain = $\frac{70}{100}\times 1.610V$ = 1.127V g...(ii)
Total mass of the solute = 0.3654V + 1.127V = 1.4924V g
Moles of $H_{2}S{O}_4$ = $\frac{\text{given mass}}{\text{molar mass}}=\frac{1.4924V}{98}$ mol
Total volume of the solution = 2V mL = $\frac{2V}{1000}$ L
Molarity = $\frac{\text{moles of solute}}{\text{volume of solution in L}}$
Molarity = $\frac{1.4924V/98}{2V/1000}$ = 7.61 M