# Chemistry Lab Calculations: Molarity and Stock Solutions

## Question 1:

How many milliliters of a 2.50 M HCl stock solution are needed to prepare 250.0 mL of a 0.100 M HCl solution?

## Question 2:

What is the molar concentration of glucose in a beaker after spilling 35.0 mL of a 3.00 M aqueous glucose solution into 150.0 mL of pure water?

## Answer 1:

10.0 mL of the 2.50 M stock solution are required to make 250.0 mL of 0.100 M HCl.

## Answer 2:

The molarity of glucose in the beaker after the spill is approximately 0.57 M.

In order to calculate the volume of the 2.50 M stock solution needed to prepare 250.0 mL of 0.100 M HCl, we can use the formula M1V1 = M2V2, where M1 is the molarity of the stock solution, V1 is the volume of the stock solution needed, M2 is the desired molarity, and V2 is the desired volume.

Substituting the given values into the formula, we get V1 = (0.100 mol/L * 250.0 mL) / 2.50 mol/L, which simplifies to 10.0 mL of the stock solution needed.

After the spill of 35.0 mL of a 3.00 M glucose solution into a beaker containing 150.0 mL of water, the total volume of the solution becomes 185.0 mL. To find the new molarity of glucose in the beaker, we calculate the amount of moles of glucose (3.00 mol/L * 0.035 L) and divide it by the total volume in liters (0.185 L), resulting in a new molarity of approximately 0.57 M.