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You start with 285mL of 0.35M solution of NaCl. Sufficient amount of water isadded so that the volume increases to 550mL. What will be concentration ofthe new solutionO none of the other answers are correct0.17 M0.18 M0.675 MO 0.10 M

User Louis Ameline
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You start with 285mL of 0.35M solution of NaCl. Sufficient amount of water is added so that the volume increases to 550mL. What will be concentration of the new solution.

Initial volume = V₁ = 285 mL

Initial concentration = M₁ = 0.35 M

Final volume = V₂ = 550 mL

Final concentation = M₂ = ?

We are only adding water to our solution, so we are diluting the solution. The number of moles of NaCl will remain constant. Since the amount of the solute is constant, in dilution exercises we can use this formula:

V₁ * M₁ = V₂ * M₂

If we solve it for the final concentration we get:

M₂ = V₁ * M₁ / V₂

If we replace by the given values we obtain:

M₂ = 285 mL * 0.35 M / 550 mL

M₂= 0.18 M

User Alban Dericbourg
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