# How To Solve Concentration Problems

Notice that it was necessary to subtract the mass of the $$\ce$$ $$\left( 150 \: \text \right)$$ from the mass of solution $$\left( 3000 \: \text \right)$$ to calculate the mass of the water that would need to be added.

Notice that it was necessary to subtract the mass of the $$\ce$$ $$\left( 150 \: \text \right)$$ from the mass of solution $$\left( 3000 \: \text \right)$$ to calculate the mass of the water that would need to be added.

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SOLUTION We can substitute the quantities given in the equation for mass/mass percent: $$\mathrm$$ Sometimes you may want to make up a particular mass of solution of a given percent by mass and need to calculate what mass of the solute to use.

Using mass percent as a conversion can be useful in this type of problem.

Using the dilution equation, we have Concentrating solutions involves removing solvent.

Usually this is done by evapourating or boiling, assuming that the heat of boiling does not affect the solute.

A dilute solution is one in which there is a small amount of solute in a given amount of solvent.

A dilute solution is a concentrated solution that has been, in essence, watered down.

Think of the frozen juice containers you buy in the grocery store.

What you have to do is take the frozen juice from inside these containers and usually empty it into 3 or 4 times the container size full of water to mix with the juice concentrate and make your container of juice.

) In both dilution and concentration, the amount of solute stays the same. Note that this equation gives only the initial and final conditions, not the amount of the change. If 25.0 m L of a 2.19 M solution are diluted to 72.8 m L, what is the final concentration?

This gives us a way to calculate what the new solution volume must be for the desired concentration of solute. Solution It does not matter which set of conditions is labelled 1 or 2, as long as the conditions are paired together properly.

## Comments How To Solve Concentration Problems

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