A 37.0 mL aliquot of a 0.600 M stock solution must be diluted to 0.100 M. Assuming the volumes are additive, how much water should be added?

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,

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

∴

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

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Explanation:

The number of moles in the concentrated stock solution is:

The number of moles in the diluted solution is:

Since the number of moles does not change when pure solvent is added:

Substituting the known values:

The volume of solvent added should be 222.0 ml - 37.0ml = 185ml

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