Question

A 5.21 mass % aqueous solution of urea (CO(NH2)2) has a density of 1.15 g/mL. Calculate the molarity of the solution. Give your answer to 2 decimal places.

Answer 1

Answer:

Molarity is 0.99 M

Explanation:

5.21% by mass, is a sort of concentration which shows the mass of solute in 100 g of solution.

Molarity is a sort of concentration that indicates the moles of solute in 1 L of solution (mol/L)

Let's find out the volume of solution by density.

Solution density = Solution mass / Solution volume

1.15 g/mL = 100 g / Solution volume

Solution volume = 100 g / 1.15 g/mL → 86.9 mL

We must have the volume of solution in L, so let's convert it.

86.9 mL / 1000 = 0.0869 L

Now, we have to determine the moles of solute (urea)

5.21 g . 1 mol / 60 g = 0.0868 moles

Mol/L = Molarity → 0.0868 moles / 0.0869L  = 0.99 M

Answer 2

Answer:

$\large \boxed{\text{1.00 mol/L}}$

Explanation:

Molar concentration = moles/litres

So, we need both the number of moles and the volume.

1. Volume

Assume a volume of 1 L.

That takes care of that.

2. Moles of urea

(a) Mass of solution

$\text{ Mass of solution} = \text{1000 mL} \times \dfrac{\text{1.15 g solution}}{\text{1 mL}} = \text{1150 g solution}$

(b) Mass of urea

$\text{Mass of urea} = \text{1150 g solution}\times \dfrac{\text{5.21 g urea}}{\text{100 g solution}} = \text{59.92 g urea}$

(c) Moles of urea

$\text{Moles of urea} = \text{59.92 g urea} \times \dfrac{\text{1 mol urea}}{\text{60.06 g urea}} = \text{1.00 mol urea}$

3. Molar concentration

$\text{Molar concentration} = \ \dfrac{\text{1.00 mol}}{\text{1 L}} = \textbf{1.00 mol/L}\\\text{The molar concentration of the urea is \large \boxed{\textbf{1.00 mol/L}} }$