225cm3of a 0.225moldm-3solution was concentration until the concentration becomes tripled. Calculate the: a. Final Concentration b. Final volume of solution. c. volume of water removed by evaporation Brainliest answer to anyone who gets it right

A. 0.675 moldm¯³

B. 75 cm³

C. 150cm³

Explanation:

Data obtained from the question include the following:

Volume of stock solution (V1) = 225 cm³

Molarity of stock solution (M1) = 0.225 moldm¯³

A. Determination of the molarity of the concentrated solution.

This is illustrated below:

Molarity of stock solution (M1) = 0.225 moldm¯³.

Molarity of concentrated solution (M2) =?

Molarity of concentrated solution (M2) = 3 x the molarity of stock solution.

Molarity of concentrated solution (M2) = 3 x 0.225 = 0.675 moldm¯³

Therefore, the molarity of the concentrated solution is 0.675 moldm¯³

B. Determination of the volume of the concentrated solution.

This is illustrated below:

Volume of stock solution (V1) = 225 cm³ = 225/1000 = 0.225 dm³

Molarity of stock solution (M1) = 0.225 moldm¯³

Molarity of concentrated solution (M2) = 0.675 moldm¯³

Volume of concentrated solution (V2) =.?

M1V1 = M2V2

0.225 x 0.225 = 0.675 x V2

Divide both side by 0.675

V2 = (0.225 x 0.225) / 0.675

V2 = 0.075 dm³

Converting 0.075 dm³ to cm³, we obtained:

0.075 x 1000 = 75 cm³

Therefore, the volume of the concentrated solution is 75 cm³.

C. Determination of the volume of water removed by evaporation.

This can be obtained by finding the difference in volume of the stock solution and concentrated solution. This is illustrated below:

Volume of stock solution (V1) = 225 cm³

Volume of the concentrated solution = 75 cm³.

Volume of water removed =..?

Volume of water removed = Volume of stock solution – Volume of concentrated solution.

Volume of water removed = V1 – V2

Volume of water removed = 225 – 75

Volume of water removed = 150 cm³

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