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Answer : The volumes of stock solution and distilled water will be, 20 mL and 80 mL respectively.
Explanation : Given,
Mass of NaOH = 60 g
Volume of stock solution = 300 mL
Molar mass of NaOH = 40 g/mol
First we have to calculate the molarity of stock solution.
[tex]\text{Molarity}=\frac{\text{Mass of }NaOH\times 1000}{\text{Molar mass of }NaOH\times \text{Volume of solution (in mL)}}[/tex]
Now put all the given values in this formula, we get:
[tex]\text{Molarity}=\frac{60g\times 1000}{40g/mole\times 300mL}=5mole/L=5M[/tex]
Now we have to determine the volume of stock solution and distilled water mixed.
Formula used :
[tex]M_1V_1=M_2V_2[/tex]
where,
[tex]M_1\text{ and }V_1[/tex] are the molarity and volume of stock solution.
[tex]M_2\text{ and }V_2[/tex] are the molarity and volume of diluted solution.
From data (A) :
[tex]M_1=5M\\V_1=20mL\\M_2=1M\\V_2=?[/tex]
Putting values in above equation, we get:
[tex]5M\times 20mL=1M\times V_2\\\\V_2=100mL[/tex]
Volume of stock solution = 20 mL
Volume of distilled water = 100 mL - 20 mL = 80 mL
From data (B) :
[tex]M_1=5M\\V_1=20mL\\M_2=1M\\V_2=?[/tex]
Putting values in above equation, we get:
[tex]5M\times 20mL=1M\times V_2\\\\V_2=100mL[/tex]
Volume of stock solution = 20 mL
Volume of distilled water = 100 mL - 20 mL = 80 mL
From data (C) :
[tex]M_1=5M\\V_1=60mL\\M_2=1M\\V_2=?[/tex]
Putting values in above equation, we get:
[tex]5M\times 60mL=1M\times V_2\\\\V_2=300mL[/tex]
Volume of stock solution = 60 mL
Volume of distilled water = 300 mL - 60 mL = 240 mL
From data (D) :
[tex]M_1=5M\\V_1=20mL\\M_2=1M\\V_2=?[/tex]
Putting values in above equation, we get:
[tex]5M\times 60mL=1M\times V_2\\\\V_2=300mL[/tex]
Volume of stock solution = 60 mL
Volume of distilled water = 300 mL - 60 mL = 240 mL
From this we conclude that, when 20 mL stock solution and 80 mL distilled water mixed then it will result in a solution that is approximately 1 M NaOH.
Hence, the volumes of stock solution and distilled water will be, 20 mL and 80 mL respectively.
When 300 mL of a solution containing 60. g of NaOH is diluted with 1200 mL of water, the resulting solution is 1 M NaOH.
60. g of NaOH is dissolved in enough distilled water to make 300 mL (V₁) of a stock solution.
We will calculate the molarity of the stock solution (M₁) using the following expression.
M₁ = mass of solute / molar mass of solute × liters of solution
M₁ = 60. g / (40.00 g/mol) × 0.300 L = 5.0 M
We want to prepare a solution with a concentration of 1 M (M₂) from a 5.0 M stock solution. We will have to carry out a dilution.
Dilution is the addition of solvent, which decreases the concentration of the solute in the solution.
We can calculate the final volume of the solution (V₂) using the dilution rule.
M₁ × V₁ = M₂ × V₂
V₂ = M₁ × V₁ / M₂ = 5.0 M × 300 mL / 1 M = 1500 mL
The volume of distilled water is the difference between the volume of the dilute solution and the volume of the stock solution.
V(H₂O) = V₂ - V₁ = 1500 mL - 300 mL = 1200 mL
When 300 mL of a solution containing 60. g of NaOH is diluted with 1200 mL of water, the resulting solution is 1 M NaOH.
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