# Question #7eb0b

##### 1 Answer

#### Answer:

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 = n/V = "1 mole"/"1 L" = "1 mol/L" = "1 M"#

In your case, the concentration of the solution is given in *milimolar*, *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 = n/V = "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

#C = n/V implies n = C * 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

#n = C * V = "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 = m/V implies m = rho * V#