Question

The density of a solution prepared by dissolving 120 g urea (Mol. Mass = 60) in 1000 g of water is 1.15 g/mL. The molarity of this solution is:
a.) 0.50 M
b.) 1.78 M
c.) 1.02 M
d.) 2.05 M

Answer 1

Hint: To solve this question first we have to understand the term molarity. Molarity is used to define the concentration of a solution.

Complete step by step answer:
- Molarity is the method to define concentration of a solution in a unit of mol/L. Molarity of any solution is defined as, “number of moles of solute per liter of solution”. It is represented by the symbol ‘M’.
- The formula of molarity is mentioned below:
\[\text{Molarity=}\dfrac{\text{moles of solute (mol)}}{\text{volume of solution (L)}}\text{x}1000\]
- We need to calculate the moles of solute and volume to calculate the molarity.
Number of moles is calculated by the ratio of given weight to the molar mass of the compound
Number of moles =\[\dfrac{120}{60} = 2\]
Volume is calculated by the ratio of mass and density. Mass of the solution will be equal to the sum of mass of water and mass of urea.
Mass of solution = $1000 + 120 = 1120g$
Given, the density of the solution = 1.15g/ml
\[volume=\dfrac{1120}{1.15}=973.9\]
Now, we can calculate the Molarity of the solution
\[\begin{align}
 & \text{Molarity=}\dfrac{\text{moles of solute (mol)}}{\text{volume of solution (L)}}\text{x}1000 \\
 & =\dfrac{2}{973.9}\times 1000 \\
 & =2.05mol/L \\
\end{align}\]
So the correct answer is “D”:

Note: Do not get confused between molarity and molality. Molality of any solution is defined as, “the number of moles of solute present per kg of solvent”. Therefore, its unit is mol/kg. It is denoted by ‘m’.