Answer:
Explanation:
The number of moles in the concentrated stock solution is:
[tex]n=C_1V_1[/tex]
The number of moles in the diluted solution is:
[tex]n=C_2V_2[/tex]
Since the number of moles does not change when pure solvent is added:
[tex]C_1V_1=C_2V_2[/tex]
Substituting the known values:
[tex]0.600M\times 37.0ml=0.100M\times V_2[/tex]
[tex]V_2=0.600M\times 37.0ml/0.100M\\ \\ V_2=222.ml[/tex]
The volume of solvent added should be 222.0 ml - 37.0ml = 185ml
The volume of water that should be added is 185.0 mL
From the question,
We are to determine the volume of water the should be added.
From the dilution formula,
We have that,
[tex]M_{1}V_{1} = M_{2}V_{2}[/tex]
M₁ is the initial molarity
V₁ is the initial volume
M₂ is the final molarity
and V₂ is the final volume
Now, we will determine the final volume of the solution
From the given information
M₁ = 0.600 M
V₁ = 37.0 mL
M₂ = 0.100 M
Putting the parameters into the formula, we get
[tex]0.600 \times 37.0 = 0.100 \times V_{2}[/tex]
∴ [tex]V_{2} = \frac{0.600 \times 37.0}{0.100}[/tex]
[tex]V_{2} = \frac{22.2}{0.100}[/tex]
V₂ = 222.0 mL
This is the final volume of the solution.
Now, for the volume of water that should be added,
Volume of water that should be added = V₂ - V₁
That is,
Volume of water that should be added = 222.0 mL - 37.0 mL
Volume of water that should be added = 185.0 mL
Hence, the volume of water that should be added is 185.0 mL
Learn more here: https://brainly.com/question/18596514
