**Surviving C _{1}V_{1 }= C_{2}V_{2}.**

Remember

C_{1}V_{1}_{ }= C_{2}V_{2}.

C_{1} = original concentration of the solution, before it
gets watered down or diluted.

C_{2} = final concentration of the solution, after dilution.

V_{1} = volume about to be diluted

V_{2} = fianl volume after dilution

By drawing the "X" through the equal sign and filling in the formula with letters of a size permitted by the borders of the "X", it reminds you that :

for all dilution problems C_{1}> C_{2, }and V_{1}< V_{2}.

It makes sense because to dilute, we add
water. This *increases* the volume but *lowers* concentration.

**Examples by Type**:

1. Easiest: Joe has a 2 g/L solution. He dilutes it and creates 3 L of a 1 g/L solution.

How much of the original solution did he dilute?

Solution: We
are looking for V_{1}:

C_{1}V_{1
}= C_{2}V_{2}

_{ }2x =
1(3)

x
= 1.5 L

2. A little trickier: Joe has 20 L of a 2 g/L solution. He diluted it, and created 3 L of a 1 g/L solution.

How did he make such a solution.

We're only going to use part of the 20 L. Remember we have
to end up with 3 L *after* dilution, so not only do we have to start with
less than 20 L but also less than 3 L. How much is unknown = V_{1}, and
it amounts to the same problem as in example 1, but don't use the 20L.

C_{1}V_{1
}= C_{2}V_{2}

_{ }2x =
1(3)

x
= 1.5 L.

3. Trickier too: Joe has 20 L of a 2 g/L solution. To this solution he adds 30 L. What is the final concentration of the solution?

If 30 L is added to the 20 L, then the volume
about to be diluted, V_{1}, is 20 L. And adding 30 L of water to a 20 L
solution creates a fianl volume, V_{2}, of 50 L. Our unknown is C_{2}.

C_{1}V_{1 }= C_{2}V_{2}

_{ }2(20) = x(20 + 30)

x = 40/50 = 0.80 g/L