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Problem #1: What is the percent by volume of a solution formed by mixing 25mL of isopropanol with 45mL water?
Problem #2: A photographic “stop bath” contains 160mL of pure acetic acid, HC2H3O2(l) in 650mL solution. What is the v/v concentration of acetic acid in the stop bath?
Problem #1: What is the percent by volume of a solution formed by mixing 25mL of isopropanol with 45mL water?
Problem #2: A photographic “stop bath” contains 160mL of pure acetic acid, HC2H3O2(l) in 650mL solution. What is the v/v concentration of acetic acid in the stop bath?
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Answer:
1) We have the following information:
% V / V (percentage of volume per volume) =?
V1 (solute volume) = 25 mL
V (volume of the solution) = solute + solvent = 25 mL + 45 mL = 70 mL
We apply the data to the formula of volume percentage of the solute per volume of solution, we will have:
\%\:v/v = \dfrac{v_1}{v} *100%v/v=
v
v
1
∗100
\%\:v/v = \dfrac{25}{70} *100%v/v=
70
25
∗100
\%\:v/v = \dfrac{2500}{70}%v/v=
70
2500
2)
=v/v x 100
=160/650× 100
=0.246×100
=24.6
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