Finding volume of a solution using %W/V

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In summary, to prepare a 6.0% W/V hydrogen peroxide solution using a 30% W/V stock solution, you will need to mix 0.1275 L of the stock solution with 297.5 mL of water for a total volume of 425 mL. The system of equations used to solve this problem is v + w = 425 and (0.30v)/(v+w) = 0.06.
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Ace.
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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?
 
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  • #2
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]\[
v + w = 425\,ml
\]
[/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] \[
\frac{{0.30v}}{{v + w}} = 0.06
\]
[/itex]

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

FAQ: Finding volume of a solution using %W/V

How do you calculate the %W/V of a solution?

The %W/V (weight/volume) of a solution is calculated by dividing the weight of the solute (in grams) by the volume of the solution (in milliliters) and multiplying by 100.

2. Why is it important to know the %W/V of a solution?

Knowing the %W/V of a solution is important because it tells you the concentration of the solute in the solution. This is important for determining the appropriate amount of a solution to use in experiments and for accurately measuring the concentration of a substance.

3. What is the difference between %W/V and other concentration units such as molarity or molality?

%W/V is a weight-to-volume concentration unit, meaning it expresses the weight of a solute in a given volume of a solution. Molarity and molality, on the other hand, are based on the number of moles of solute in a given volume or mass of solvent, respectively.

4. Can the %W/V of a solution change?

Yes, the %W/V of a solution can change if the weight or volume of the solute or solvent changes. It can also change if the concentration of the solute changes, for example, if more solute is added to the same volume of solution.

5. How can you measure the volume of a solution to determine its %W/V?

The volume of a solution can be measured using a graduated cylinder, a volumetric flask, or a pipette. The weight of the solute can be measured using a balance. Once you have these measurements, you can plug them into the %W/V formula to calculate the concentration of the solution.

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