What volume of 2.5% (m/v) koh can be prepared from 125 ml of a 5.0% (m/v) koh solution?

Question

What volume of 2.5% (m/v) koh can be prepared from 125 ml of a 5.0% (m/v) koh solution?

in progress 0
2022-05-23T14:18:12+00:00 2 Answers 0

Answers ( 2 )

  1. Alexandra
    0
    2022-05-23T14:19:27+00:00

    Answer: 250 mL

    Explanation: %(m/v) stands for mass by volume percentage. 2.5%(m/v) means 2.5 grams of a solute present in 100 mL of a solution.

    The question asks to calculate the volume of 2.5%(m/v) KOH solution that can be prepared from 125 mL of a 5.0%(m/v) KOH solution.

    It’s a dilution problem and could easily be solved by using dilution equation:

    [tex]C_1V_1=C_2V_2[/tex]

    where [tex]C_1[/tex] is the concentration before dilution and [tex]C_2[/tex] is the concentration after dilution. Similarly, [tex]V_1[/tex] is the volume before dilution and [tex]V_2[/tex] is the volume after dilution.

    Let’s plug in the values in the equation:

    [tex]5.0(125mL)=2.5(V_2)[/tex]

    [tex]V_2=frac{5.0(125mL)}{2.5}[/tex]

    [tex]V_2[/tex] = 250 mL

    So, 250 mL of 2.5%(m/v) KOH solution can be prepared from 125 mL of 5.0%(m/v) KOH solution.

  2. Josie97
    0
    2022-05-23T14:19:35+00:00

    the volume of 2.5% m/v koh which can prepared from 125 ml of a 5% koh solution is calculated using the following formula

    m1v1= m2 v2

    M1= 5/100= 0.05
    v1= 125
    m2=2.5/100=0.025
    V2=?
    v2= m1v1/m2

    =0.05 x125 /0.025=250 ml

Leave an answer

Browse
Browse

45:5+15*4 = ? ( )