What volume of 2.5% (m/v) KOH can be prepared from 125 mL of a 5.0% KOH solution?
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Question “What volume of 2.5% (m/v) KOH can be prepared from 125 mL of a 5.0% KOH solution?”
What volume of 2.5% (m/v) KOH can be prepared from 125 mL of a 5.0% KOH solution?
Answer
Explanation:
The
Short answer
The volume of the target solution should be equal
You can think of it this way: The concentration of your target solution should be half less than the concentration in the stock solution.
You’re retaining the amount
Solute
unchanged can be used to indicate that doubling the volume will yield half as the initial concentration
Diluting
The initial solution was by a
Factor
Of
The long answer is now available.
You’re now dealing with
A
Volume vs mass
Percent concentration
mass of solute is the definition of this term, which is often expressed in
Grams
Divided by volume of solution, often expressed in
milliliters
Multiplied by
color(blue )(“%m/v) = “mass solute”/”volume solution” xx 100. %m/v=mass solutevolume solutionx100
The stock solution for your case will contain
“5.0% g/mL” = m_(KOH)/”125 mL” xx 1005.0% g/mL=mKOH125 mLx100
m_(KOH) = (5.0″g”/color(red)(cancel(color(black)(“mL”))) * 125 color(red)(cancel(color(black)(“mL”))))/100 = “6.25 g”
mKOH
=
5.0
g
mL
125
mL
100
=
6.25 g
Your stock solution should contain these things
This means you are now a “certified” user.
“2.5% g/mL” = “6.25 g”/V_”sol” * 1002.5% g/mL=6.25 gVsol100
V_”sol” = (6.25 color(red)(cancel(color(black)(“g”))))/(2.5color(red)(cancel(color(black)(“g”)))/”mL”) * 100 = color(green)(“250 mL”)
Vsol
=
6.25
g
2.5
g
mL
100
=
250 mL
Conclusion
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