Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store
seo-qna
SearchIcon
banner

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
VerifiedVerified
482.1k+ views
1 likes
like imagedislike image
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:
Molarity=moles of solute (mol)volume of solution (L)x1000
- 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 =12060=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=11201.15=973.9
Now, we can calculate the Molarity of the solution
Molarity=moles of solute (mol)volume of solution (L)x1000=2973.9×1000=2.05mol/L
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’.
Latest Vedantu courses for you
Grade 8 | CBSE | SCHOOL | English
Vedantu 8 CBSE Pro Course - (2025-26)
calendar iconAcademic year 2025-26
language iconENGLISH
book iconUnlimited access till final school exam
tick
School Full course for CBSE students
EnglishEnglish
MathsMaths
ScienceScience
₹49,800 (9% Off)
₹45,300 per year
Select and buy