Concentration of solutions [closed]








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I'm stuck with this problem. If I have 200 grams of a solution at 30% how much water should I add so that the concentration becomes 25%?




The answer is that for a simple dilution the following formula applies:


$$c\_1m\_1 = c\_2m\_2$$ $$ m\_2 = \frac{c\_1m\_1}{c\_2} = \frac{(200g)(30\text{%})}{20\text{%}} = 240g$$


Therefore the mass to add is $(240g - 200g) = 40g$ of $\ce{H2O}$ (which is 40 ml of $\ce{H2O}$). 


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There is a rule for. In german wikipedia there is a article for 'Mischkreuz', I don't know the equivalent in english, maybe the 'cross over rule'. 


If you have two solutions with different concentrations and have to mix up a concentration in the middle. The difference between the start concentrations and the final concentration gives the volume ratio of the input solutions. 


In your example: 


You need 25 parts of your concentrated solution. (difference: 25% - 0%) 


And 5 parts of water. (difference: 30% - 25%) 


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