Question 838678
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Let W be the volume of water (in mL) to add to your original solution.


When you add, your final solution will have the volume of (300+W) mL. 


This volume will contain the same 0.14*300 mL of bleach as the original solution.


So, the concentration of the final solution will be this ratio  {{{(0.14*300)/(300+W)}}} of the "pure bleach" to the total liquid volume.


According to the condition, this ratio must be equal to 10%, or 0.1.

It gives you an equation

{{{(0.14*300)/(300+W)}}} = 0.1.


It is your major/basic equation for the given problem.

As soon as you got this equation, the setup part is done and completed.

This equation is often called a "concentration" equation, since its both sides describe the concentration of the final solution.


To solve the equation, multiply both sides by (300+W). You will get

0.14*300 = 0.1*(300 + W),

0.14*300 = 0.1*300 + 0.1*W,

0.14*300 - 0.1*300 = 0.1*W  ====>  0.04*300 = 0.1*W  ====>  W = {{{(0.04*300)/0.1}}} = 0.4*300 = 120.


<U>Answer</U>.  You need to add 120 mL of water.


<U>Check</U>.   {{{(0.14*300)/(300+12))}}} = 0.1 = 10%.   ! Correct !
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