Question

If I add 375 of water to 125 mL of a 0.90 M NaOH solution, what will the molarity of the diluted solution be?

Answer 1

Answer:

The molarity of the diluted solution of NaOH is 0.225 M NaOH

Explanation:

Here, we have, initial concentration of the solution = 0.90 M NaOH

Volume of initial solution = 125 mL

Volume of added water = 375 mL

Total volume of diluted solution = 375 mL + 125 mL = 500 mL

Number of moles in the diluted solution = 0.9 × 0.125 = 0.1125 moles of NaOH

Derived from the following relation;

0.9 M NaOH indicates that 1000 mL contains 0.9 moles of NaOH

Therefore, 1 mL contains 0.9/1000 moles of NaOH and 125 mL contains 0.9/1000×125 or 0.9 × 0.125 moles of NaOH

Therefore, the diluted 500 mL solution contains 0.1125 moles of NaOH. Therefore, 1000 mL of the diluted solution will contain $1000 \times \frac{0.1125}{500} \ moles$ of NaOH or 0.225 moles of NaOH

Hence the molarity of the diluted solution of NaOH, which is the number of moles per 1000 mL is 0.225 M.

Answer 2

Answer:

0.225M

Explanation:

Step 1:

Data obtained from the question. This includes:

Volume of the stock solution (V1) = 125mL

Molarity of the stock solution (M1) = 0.90 M

Volume of the diluted solution (V2) = 125 + 375 = 500mL

Molarity of the diluted solution (M2) =..?

Step 2:

Determination of the molarity of the diluted solution.

This is obtained by using the dilution formula as follow:

M1V1 = M2V2

0.9 x 125 = M2 x 500

Divide both side by 500

M2 = (0.9 x 125) /500

M2 = 0.225M

Therefore, the molarity of the diluted solution is 0.225M