# You have 67.0 mL of a 0.400 M stock solution that must be diluted to 0.100 M. Assuming the volumes are additive, how much water should you add?

The thing to remember about dilutions is that the ratio that exists between the concentration of the concentrated solution and that of the diluted solution is equal to the ratio that exists between the volume of the diluted solution and that of the concentrated solution.

This ratio is called the dilution factor. You thus have

and

and therefore

Assuming that the volumes are addictive, you will have to add

The answer is rounded to three sig figs.

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To dilute the 67.0 mL of 0.400 M stock solution to 0.100 M, you can use the formula:

(C_1V_1 = C_2V_2),

where (C_1) and (V_1) are the concentration and volume of the initial solution, and (C_2) and (V_2) are the concentration and volume of the final solution.

Given: (C_1 = 0.400) M, (V_1 = 67.0) mL, (C_2 = 0.100) M, (V_2) (unknown).

Rearranging the formula to solve for (V_2):

(V_2 = \frac{{C_1 \cdot V_1}}{{C_2}})

(V_2 = \frac{{0.400 , \text{M} \cdot 67.0 , \text{mL}}}{{0.100 , \text{M}}})

(V_2 = \frac{{26.8 , \text{mL}^2 \cdot \text{M}}}{{0.100 , \text{M}}})

(V_2 = 268 , \text{mL}).

So, you would need to add (268 , \text{mL} - 67 , \text{mL} = 201 , \text{mL}) of water.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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