Answer:
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
