Question

A solution is prepared by taking 25.0 mL of a stock solution of NaOH and diluting it to a final volume of 350. mL. the molarity of the diluted solution is 0.042 m. Which of the following options correctly describe these solutions? select all that apply.

a. The portion of stock solution used contained 0.0147 moles of NaOH. b. The stock solution has a molarity of 0.59M. c. The stock solution has a molarity of 0.030M. d. The 350.mL of diluted solution contains 0.042 moles of NaOH

Answer 1

Step-by-step explanation:

The amount of NaOH in moles present in the diluted solution can be calculated using the formula:

moles of solute = Molarity × Volume of solution (in liters)

Here, the final volume of the diluted solution is 350 mL = 0.35 L, and the molarity of the diluted solution is 0.042 M. Therefore, the moles of NaOH in the diluted solution are:

moles of NaOH = 0.042 M × 0.35 L = 0.0147 moles

This means that option a is correct.

To calculate the molarity of the stock solution, we can use the formula:

M1V1 = M2V2

where M1 and V1 are the molarity and volume of the stock solution, and M2 and V2 are the molarity and volume of the diluted solution. Rearranging this formula, we get:

M1 = M2 × V2 / V1

Plugging in the given values, we get:

M1 = 0.042 M × 0.35 L / 0.025 L = 0.588 M

Therefore, option b is correct.

Option c is not correct, as we just calculated that the molarity of the stock solution is 0.588 M, not 0.030 M.

Finally, option d is correct, as we calculated earlier that the diluted solution contains 0.0147 moles of NaOH.