Question

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

Answer 1

     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

Answer 2

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:

$C_1V_1=C_2V_2$

where $C_1$ is the concentration before dilution and $C_2$ is the concentration after dilution. Similarly, $V_1$ is the volume before dilution and $V_2$ is the volume after dilution.

Let's plug in the values in the equation:

$5.0(125mL)=2.5(V_2)$

$V_2=\frac{5.0(125mL)}{2.5}$

$V_2$ = 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.