Question

Dissolving 120 g of urea (mol. wt. 60) in 1000 g of water gave a solution of density 1.15 g/mL . The molarity of the solution is:
A: 1.78 M
B: 2.0 M
C: 2.05 M
D: 2.22 M

Answer 1

Hint: Here concept of molarity is used. Molarity is defined as the number of moles of solute that is present in 1 L of solution. So,
$Molarity(M) = {\dfrac{Moles\;of\;solute}{Volume\;of\;solution\;in\;litre}}$

Complete step by step answer:
Thus, in order to calculate a solution's molarity we should know the number of moles of solute that is present in 1 L i.e. 1000 mL of solution.
We will use the molar mass of urea to determine the number of moles present in the sample which is given as,
Mass of urea in solution = 120 g
Molar mass of urea = 60 g
${\text{Number of moles = }}\dfrac{{{\text{Mass of urea in solution}}}}{{{\text{Molecular mass of urea}}}} = \dfrac{{120}}{{60}} = 2$moles of urea
Now, we already know that the solution comprises 120 g of urea (solute) and 1000 g of water (solvent). So, the total mass of the solution becomes:
Mass of solution = Mass of solute + Mass of solvent
i.e. ${\text{Mass of solution}} = {\text{ 120 g + 1000 g = 1120 g}}$
From the question, we also know that the solution has a density of $1.15\; g mL^{−1}$, thus it means that each 1 mL of solution possesses a mass of 1.15 g.
From the solution’s density, we can calculate its volume:
$Volume = \dfrac{{Mass}}{{Density}} = \dfrac{{1120}}{{1.15}} = 973.9\;mL$
Now, our ultimate aim is to find out the number of moles of solute present in 1000 mL of solution. We can utilise the known composition of the solution as a conversion factor to get number of moles of solute:
$1000 \times \dfrac{2}{{973.9}} = 2.0535{\text{ moles of urea}}$

Therefore, the molarity of the solution is equal to 2.05 M (Option C).

Note: Don’t get confused between molarity and molality. Molarity is the number of moles of solute per volume of solution (litres) while Molality is the number of moles of solute per weight of solvent (kilogram). These two have a relationship as below: