All Courses
Explore Courses Class 12 Exam Class 12 Test Series Free Notes EduRev Infinity Get the App Login Join for Free
Sign in Open App
Class 12 Exam  >  Class 12 Questions  >  Can you explain the answer of this question b... Start Learning for Free
Can you explain the answer of this question below:
 For an aqueous solution, freezing point is _0.186ºC . Boiling point of the same solution is
(Kƒ = 1.86º K mol_1 kg) and
Kb = 0.512º K mol_1 kg)                                                 [AIEEE-2002]
  • A:
    0.186º 
  • B:
    100.0512º 
  • C:
    1.86º 
  • D:
    5.12º
The answer is B.
Clear all your Class 12 doubts with EduRev
Verified Answer
Can you explain the answer of this question below:For an aqueous solut...
ΔTb/Kb = ΔTf/Kf
0.186/18.6 = ΔTb/0.512
ΔTb = 0.1x0.512
= 0.0512.
Boiling point = boiling point (of water) + ΔTb
=> 100 + 0.0512
= 100.0512
View all questions of this test
Most Upvoted Answer
Can you explain the answer of this question below:For an aqueous solut...
Free Test
FREE
Start Free Test
Start Free Test
Community Answer
Can you explain the answer of this question below:For an aqueous solut...
Explanation:

Given Data:
- Freezing point depression constant (Kf) = 1.86 °C kg/mol
- Boiling point elevation constant (Kb) = 0.512 °C kg/mol
- Freezing point depression = 0.186 °C

Calculating Molality:
- Molality (m) = ΔTf / Kf
- Molality (m) = 0.186 / 1.86 = 0.1 mol/kg

Calculating Boiling Point Elevation:
- ΔTb = Kb * m
- ΔTb = 0.512 * 0.1 = 0.0512 °C

Calculating Boiling Point:
- Boiling Point (Tb) = Normal Boiling Point + ΔTb
- Boiling Point (Tb) = 100 + 0.0512 = 100.0512 °C
Therefore, the boiling point of the solution is 100.0512 °C, which corresponds to option B.
Explore Courses for Class 12 exam

Similar Class 12 Doubts

Direction: Read the passage given below and answer the following questions:Boiling point or freezing point of liquid solution would be affected by the dissolved solids in the liquid phase. A soluble solid in solution has the effect of raising its boiling point and depressing its freezing point. The addition of non-volatile substances to a solvent decreases the vapour pressure and the added solute particles affect the formation of pure solvent crystals.According to many researches the decrease in freezing point directly correlated to the concentration of solutes dissolved in the solvent. This phenomenon is expressed as freezing point depression and it is useful for several applications such as freeze concentration of liquid food and to find the molar mass of an unknown solute in the solution. Freeze concentration is a high quality liquid food concentration method where water is removed by forming ice crystals. This is done by cooling the liquid food below the freezing point of the solution. The freezing point depression is referred as a colligative property and it is proportional to the molar concentration of the solution (m), along with vapour pressure lowering, boiling point elevation, and osmotic pressure. These are physical characteristics of solutions that depend only on the identity of the solvent and the concentration of the solute. The characters are not depending on the solute’s identity.Q.Colligative properties are

Question Description
Can you explain the answer of this question below:For an aqueous solution, freezing point is _0.186ºC . Boiling point of the same solution is(Kƒ= 1.86º K mol_1kg) andKb= 0.512º K mol_1kg) [AIEEE-2002]A:0.186ºB:100.0512ºC:1.86ºD:5.12ºThe answer is B. for Class 12 2025 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about Can you explain the answer of this question below:For an aqueous solution, freezing point is _0.186ºC . Boiling point of the same solution is(Kƒ= 1.86º K mol_1kg) andKb= 0.512º K mol_1kg) [AIEEE-2002]A:0.186ºB:100.0512ºC:1.86ºD:5.12ºThe answer is B. covers all topics & solutions for Class 12 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Can you explain the answer of this question below:For an aqueous solution, freezing point is _0.186ºC . Boiling point of the same solution is(Kƒ= 1.86º K mol_1kg) andKb= 0.512º K mol_1kg) [AIEEE-2002]A:0.186ºB:100.0512ºC:1.86ºD:5.12ºThe answer is B..
Solutions for Can you explain the answer of this question below:For an aqueous solution, freezing point is _0.186ºC . Boiling point of the same solution is(Kƒ= 1.86º K mol_1kg) andKb= 0.512º K mol_1kg) [AIEEE-2002]A:0.186ºB:100.0512ºC:1.86ºD:5.12ºThe answer is B. in English & in Hindi are available as part of our courses for Class 12. Download more important topics, notes, lectures and mock test series for Class 12 Exam by signing up for free.
Here you can find the meaning of Can you explain the answer of this question below:For an aqueous solution, freezing point is _0.186ºC . Boiling point of the same solution is(Kƒ= 1.86º K mol_1kg) andKb= 0.512º K mol_1kg) [AIEEE-2002]A:0.186ºB:100.0512ºC:1.86ºD:5.12ºThe answer is B. defined & explained in the simplest way possible. Besides giving the explanation of Can you explain the answer of this question below:For an aqueous solution, freezing point is _0.186ºC . Boiling point of the same solution is(Kƒ= 1.86º K mol_1kg) andKb= 0.512º K mol_1kg) [AIEEE-2002]A:0.186ºB:100.0512ºC:1.86ºD:5.12ºThe answer is B., a detailed solution for Can you explain the answer of this question below:For an aqueous solution, freezing point is _0.186ºC . Boiling point of the same solution is(Kƒ= 1.86º K mol_1kg) andKb= 0.512º K mol_1kg) [AIEEE-2002]A:0.186ºB:100.0512ºC:1.86ºD:5.12ºThe answer is B. has been provided alongside types of Can you explain the answer of this question below:For an aqueous solution, freezing point is _0.186ºC . Boiling point of the same solution is(Kƒ= 1.86º K mol_1kg) andKb= 0.512º K mol_1kg) [AIEEE-2002]A:0.186ºB:100.0512ºC:1.86ºD:5.12ºThe answer is B. theory, EduRev gives you an ample number of questions to practice Can you explain the answer of this question below:For an aqueous solution, freezing point is _0.186ºC . Boiling point of the same solution is(Kƒ= 1.86º K mol_1kg) andKb= 0.512º K mol_1kg) [AIEEE-2002]A:0.186ºB:100.0512ºC:1.86ºD:5.12ºThe answer is B. tests, examples and also practice Class 12 tests.
Explore Courses for Class 12 exam
SaveJoin the Discussion on the App for FREE
Signup to solve all Doubts
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev
Get App
© EduRev
Education Revolution
Signup to see your scores go up
within 7 days!
Takes less than 10 seconds to signup