# Question 7eb0b

Sep 12, 2015

Use the definition of molarity.

#### Explanation:

From what I can tell, you actually provided the same piece of information twice in your question.

Molarity is defined as moles of solute per liters of solution and is given in moles per liter or molar.

If you have 1 mole of a solute dissolved in 1 liter of solution, you have

$C = \frac{n}{V} = \text{1 mole"/"1 L" = "1 mol/L" = "1 M}$

In your case, the concentration of the solution is given in milimolar, $\text{mM}$, which is equivalent to milimoles per liter.

This time, if you have 1 mole of solute in 1 liter of solution, you have

1color(red)(cancel(color(black)("mole"))) * "1000 mmoles"/(1color(red)(cancel(color(black)("mole")))) = "1000 mmoles"

One mole contains 1000 mmoles, which means that the molarity of the solution can also be given as

$C = \frac{n}{V} = \text{1000 mmoles"/"1 L" = "1000 mmoles/L" = "1000 mM}$

Now, when you know the molarity of a solution and its volume, you can rearrange the above equation to solve for $n$, the number of moles of solute.

$C = \frac{n}{V} \implies n = C \cdot V$

Once you have thenumber of moles of solute, you can use its molar mass to determine what mass was initially dissolved to produce the respective solution.

So, for example, you know that you have $\text{150 mL}$ of a $\text{2000 mM}$ sucrose solution, and want to figure out how much sucrose you dissolved to make this solution.

$n = C \cdot V = \text{2000 mM" * 150 * 10^(-3)"L" = "300 mmoles}$

This is equivalent to

300color(red)(cancel(color(black)("mmoles"))) * "1 mole"/(1000color(red)(cancel(color(black)("mmoles")))) = "0.30 moles sucrose"

Now use sucrose's molar mass to get the mass of the dissolved sample

0.30color(red)(cancel(color(black)("moles"))) * "342.3 g"/(1color(red)(cancel(color(black)("mole")))) = "103 g sucrose"#

If yo uwant the mass of the entire solution, you need to use its density and its volume.

$\rho = \frac{m}{V} \implies m = \rho \cdot V$