# Question #bce3f

May 9, 2017

${V}_{c o n c} = 8.1 m L$

#### Explanation:

In a laboratory setting, it is common to store a highly concentrated solution in a stock room so that it may be used to prepare more dilute solutions needed for experiments.

By doing this, the lab can save space in the stock room, but it requires the chemist to be able to calculate the volume needed from the stock solution for the given experiment.

In order to calculate the necessary volume from the concentrated stock solution, we should first figure out how many moles of the reactant are needed for the experiment:

For an experiment involving 200.0 mL of 0.600 M $N {H}_{3}$, we need:

$200.0 m L \cdot \frac{1 L}{1000 m L} \cdot \frac{0.600 m o l}{1 L} = 0.12 m o l N {H}_{3}$

Now, if the stock solution has a concentration of 14.8 M $N {H}_{3}$, how much do we need to have 0.12 mol of $N {H}_{3}$?

$0.12 m o l N {H}_{3} \cdot \frac{1 L}{14.8 m o l N {H}_{3}} = 0.0081 L = 8.1 m L$

A short cut for this is to say:

${M}_{c o n c} \cdot {V}_{c o n c} = {M}_{\exp} \cdot {V}_{\exp}$

where ${M}_{c o n c}$ and ${V}_{c o n c}$ are the molarity and volume of the concentrated solution, and ${M}_{\exp}$ and ${V}_{\exp}$ are the molarity and volume needed for the experiment.

This way, you can say the volume needed from the concentrated solution is:

${V}_{c o n c} = \frac{{M}_{\exp} \cdot {V}_{\exp}}{M} _ \left(c o n c\right)$

${V}_{c o n c} = \frac{0.600 M \cdot 200 m L}{14.8 M}$

${V}_{c o n c} = 8.1 m L$