# What is the molarity of an aqueous solution containing 22.5 grams of glucose in 25.5 mL of solution?

Jun 18, 2017

${\text{4.90 mol L}}^{- 1}$

#### Explanation:

Start by calculating the number of moles of glucose present in $\text{25.5 mL}$ of solution.

To do that, use the molar mass of glucose

22.5 color(red)(cancel(color(black)("g"))) * "1 mole glucose"/(180.156color(red)(cancel(color(black)("g")))) = "0.1249 moles glucose"

Now, in order to find the molarity of the solution, you need to determine the number of moles of solute present in $\text{1 L} = {10}^{3}$ $\text{mL}$ of solution.

The fact that you're dealing with a solution, which, as you know, is a heterogeneous mixture, i.e. it has the same composition throughout, you can use the known composition as a conversion factor to find the number of moles of solute present in ${10}^{3}$ $\text{mL}$ of solution.

10^3 color(red)(cancel(color(black)("mL solution"))) * "0.1249 moles glucose"/(25.5color(red)(cancel(color(black)("mL solution")))) = "4.898 moles glucose"

You can thus say that the molarity of the solution is equal to

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{\text{molarity = 4.90 mol L}}^{- 1}}}}$

The answer is rounded to three sig figs.