What is the molarity of a 250 mL H2SO4 solution that was made from 20 mL of a 10 M stock solution?

There are two ways to solve this question. 

Since the sulfuric acid solution is made from a stock solution, the number of moles of sulfuric acid remains the same. In other words, the number of moles of sulfuric acid in the 20 mL, 10 M stock solution is same as those in the 250 mL of solution. Since molarity is the ratio of moles of solute and the volume of the solvent,

number of moles...

There are two ways to solve this question. 

Since the sulfuric acid solution is made from a stock solution, the number of moles of sulfuric acid remains the same. In other words, the number of moles of sulfuric acid in the 20 mL, 10 M stock solution is same as those in the 250 mL of solution. Since molarity is the ratio of moles of solute and the volume of the solvent,

number of moles in 20 mL, 10 M stock = molarity x volume.

First we need to convert 20 mL to liters (L):

20 mL x ( 1 L / 1000 mL) = .02 L

 = 10 M x .02 L moles = 0.2 moles H2SO4

These 0.2 moles of sulfuric acid are also present in the 250 mL of solution.

To solve for the concentration of the resultant solution, we must first convert 250 mL to L.

250 mL x (1 L / 1000 mL) = .25 L

concentration (M) = moles/volume = 0.2/(.25) = 0.8 M.

We can also use the equation:

C1V1 = C2V2

to solve this numerical, since this is a simple dilution scenario (stock solution is diluted by adding more solvent). Here C1 and C2 are concentrations of the solutions and V1 and V2 are their volumes.

Thus, 10 M x .02 L = C2 x .25 L

solving this, we get C2 = 0.8 M.

Therefore, option A is correct.

Hope this helps.