Answer:
To prepare 24 mL of 0.050 M NaOH solution, you would add 4 mL of 0.300 M NaOH stock solution and 20 mL of 0.300 M NaCl solution.
Explanation:
Molarity of the NaOH solution = [tex]M_1=0.300 M[/tex]
Volume of the NaOH solution = [tex]V_1=?[/tex]
Molarity of the NaOH solution after dilution= [tex]M_2=0.050 M[/tex]
Volume of the NaOH solution after dilution= [tex]V_2=24 mmL[/tex]
[tex]M_1V_1=M_2V_2[/tex] (Dilution )
[tex]V_1=\frac{M_2V_2}{M_1}=\frac{0.050 M\times 24 mL}{0.300 M}=4 mL[/tex]
Volume of NaCl solution of 0.300 M = 24 mL - 4 mL = 20 mL
To prepare 24 mL of 0.050 M NaOH solution, you would add 4 mL of 0.300 M NaOH stock solution and 20 mL of 0.300 M NaCl solution.
To prepare 24 mL of 0.050 M NaOH solution, you would add 4 mL of 0.300 M NaOH stock solution and 20 mL of 0.300 M NaCl solution
•Molarity of stock solution (M₁) = 0.3 M
•Molarity of diluted solution (M₂) = 0.05 M
•Volume of diluted solution (V₂) = 24 mL
•Volume of stock solution needed (V₁) =?
Using the dilution formula, the volume of the stock solution needed can be obtained as follow:
M₁V₁ = M₂V₂
0.3 × V₁ = 0.05 × 24
0.3 × V₁ = 1.2
Divide both side by 0.3
V₁ = 1.2 / 0.3
V₁ = 4 mL
•Volume of NaOH needed = 4 mL
•Volume of diluted solution of NaOH = 24 mL
•Volume of NaCl needed =?
Volume of NaCl needed = (Volume of diluted solution of Na) – (Volume of NaOH needed)
Volume of NaCl needed = 24 – 4
Volume of NaCl needed = 20 mL
Therefore, we can conclude as follow:
To prepare 24 mL of 0.050 M NaOH solution, you would add 4 mL of 0.300 M NaOH stock solution and 20 mL of 0.300 M NaCl solution.
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