# Dissolving 120g of urea (mol.wt 60) in 1000g of water gave a solution of density 1.15 g/mL. What is the molarity of the solution?

Jun 14, 2017

${\text{2.1 mol L}}^{- 1}$

#### Explanation:

The idea here is that a solution's molarity tells you the number of moles of solute present in $\text{1 L}$ of solution.

So in order to calculate a solution's molarity, you essentially need to know the number of moles of solute present in exactly $\text{1 L} = {10}^{3}$ $\text{mL}$ of solution.

Start by using the molar mass of urea to calculate the number of moles present in your sample

120 color(red)(cancel(color(black)("g"))) * "1 mole urea"/(60color(red)(cancel(color(black)("g")))) = "2 moles urea"

Now, you know that your solution contains $\text{120 g}$ of urea, the solute, and $\text{1000 g}$ of water, the solvent. This implies that the total mass of the solution

$\text{mass solution = mass solute + mass solvent}$

will be equal to

$\text{mass solution" = "120 g + 1000 g" = "1120 g}$

You also know that this solution has a density of ${\text{1.15 g mL}}^{- 1}$, which means that every $\text{1 mL}$ of solution has a mass of $\text{1.15 g}$.

Use the density of the solution to calculate its volume

1120 color(red)(cancel(color(black)("g"))) * "1 mL"/(1.15color(red)(cancel(color(black)("g"))) ) = "973.9 mL"

Now, your goal is to figure out the number of moles of solute present in ${10}^{3}$ $\text{mL}$ of solution, so use the known composition of the solution as a conversion factor to get

10^3 color(red)(cancel(color(black)("mL solution"))) * "2 moles urea"/(973.9color(red)(cancel(color(black)("mL solution")))) = "2.0536 moles urea"

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

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{\text{molarity = 2.1 mol L}}^{- 1}}}}$

I'll leave the answer rounded to two sig figs.