Question

Find mass % of $10m$ of aqueous solution of urea. If molar mass of urea is $60g/mol$ and density of solution is $1.2gc{m^{ - 3}}$.

Answer 1

We know that the mass percentage is defined as the mass of the solute (element) divided by the mass of the solution. Using the concept of molality, molar mass and mass percentage we will proceed to solution.

Complete step by step answer:
Given data contains:
Molality of urea solution = $10m$
Molar mass of urea = $60g/mol$
Density of the solution = $1.2g/c{m^3}$
To find: mass percentage
We know that molality is a measure of the total moles of a solute in one kilogram of a solvent.
According to the definition of molality,
$1000g$ of water contains $10{\text{moles}}$ of urea
$10{\text{moles}}$ of urea = $10 \times {\text{molar mass of urea}}$
= $10 \times 60$
=$600g$ of urea
Here we have the mass of the solute = $600g$
Now, the mass of solution will be = mass of solute + mass of solvent
= $600 + 1000$
= $1600g$
The mass percentage $= \dfrac{{{\text{mass of solute}}}}{{{\text{mass of solution}}}} \times 100$
On substituting the known values we get,
$\Rightarrow {\text{Mass percentage = }}\dfrac{{600}}{{1600}} \times 100$
On simplifying we get,
${\text{Mass percentage}} = 37.5\%$
Here, density is not necessary as we already have the measuring factor which is the molar mass of urea.

Hence, the mass percentage of $10$ moles of aqueous solution of urea is $37.5\%$.

Note: We must remember that the terms like molarity, molality, molar mass and mass percentage play an important role in determining the molar weightage of elements, reactants and products in any reaction. Molarity and molality can be quite confusing. The difference between the two is that, in molarity we measure moles of a solute in one litre of solution and in molality we measure one moles of solute in one kilogram of solution. Density is related to volume. Density is indirectly proportional to volume, which implies when volume increases, the density decreases.