Finding volume of a solution using %W/V

  • Thread starter Thread starter Ace.
  • Start date Start date
  • Tags Tags
    Volume
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
1 reply · 4K views
Ace.
Messages
52
Reaction score
0

Homework Statement



What volume of a 30% W/V hydrogen peroxide solution is required to prepare 425 mL of a 6.0% W/V solution?

Homework Equations



%w/v = (weight solute/volume solution) x 100%

The Attempt at a Solution



I'm not quite sure what to do because here I am given 2 percentages?
%w/v = (weight solute/volume solution) x 100%
Weight solute = (%w/v × volume solution)/100%
= (30% × 0.425 L)/100%​
= 0.1275 L​

Where to go from here and to apply this to the 6% solution?
 
Physics news on Phys.org
Ace. said:

Homework Statement



What volume of a 30% W/V hydrogen peroxide solution is required to prepare 425 mL of a 6.0% W/V solution?

Homework Equations



%w/v = (weight solute/volume solution) x 100%

The Attempt at a Solution



I'm not quite sure what to do because here I am given 2 percentages?
%w/v = (weight solute/volume solution) x 100%
Weight solute = (%w/v × volume solution)/100%
= (30% × 0.425 L)/100%​
= 0.1275 L​

Where to go from here and to apply this to the 6% solution?

Biggest part of solving is to create the system of equations.

Let v = volume of stock solution, the 30%.
Let w = volume of just the water for dilution, 0% solute.
You want total resulting solution 425 ml.
425 = w + v.
[itex]\[<br /> v + w = 425\,ml<br /> \][/itex]

You are taking v ml. of stock 30% and diluting it to get 6%. Using decimal fractions instead of percent, you setup a ratio relationship:
((0.30)v)/(v+w) = 0.06
[itex]\[<br /> \frac{{0.30v}}{{v + w}} = 0.06<br /> \][/itex]

Two unknown numbers and two equations. Solve.
 
Last edited: