I need help with the attached numerical on concentration of chemicals.

From the numerical we can see that a stock solution is diluted to make a solution of lesser strength. In other words, a certain volume of 12 M concentrated hydrochloric acid is used to prepare 1500 ml of 3.5 M acid solution. 

This means that the number of moles of acid is the same in both of the solutions.

Since molarity is the ratio of moles of a solute to the volume of solvent in liters, we can solve for the number of moles of acid. First, we need to convert 1500 ml to liters.

1500 ml x (1 L / 1000 ml) = 1.5 L

Moles of hydrochloric acid = molarity x volume = 3.5 M x 1.5 L = 5.25 moles.

For the concentrated solution,

volume = number of moles/molarity = 5.25 moles/12 M = 0.4375 L = 437.5 ml.

Thus, 437.5 ml of a concentrated hydrochloric acid solution (12 M) is needed to make a 1500 ml, 3.5 M solution.

Among the given options, option C (440 ml) is the closest one and hence is the answer.

Hope this helps.