The freezing point of an aqueous solution is $-0.186^oC$. If the molal elevation constant and molal depression constant of the solvent are $0.512$ and $1.86$ respectively,then the elevation in boiling point is .......... $^oC$.

  • A
    $0.186$
  • B
    $0.512$
  • C
    $0.86$
  • D
    $0.0512$

Explore More

Similar Questions

$A$ solution containing $1.8 \ g$ of a compound (empirical formula $CH_2O$) in $40 \ g$ of water is observed to freeze at $-0.465 \ ^oC$. The molecular formula of the compound is ($K_f$ of water $= 1.86 \ K \ kg \ mol^{-1}$)

If the mole fraction of the solvent decreases while preparing a solution,then ...........

If a $1 \ m$ solution of benzoic acid in benzene has a freezing point depression of $2.56 \ ^{\circ}C$ $(K_f = 5.12 \ ^{\circ}C \ kg \ mol^{-1})$ and a boiling point elevation of $2.53 \ ^{\circ}C$ $(K_b = 2.53 \ ^{\circ}C \ kg \ mol^{-1})$,select the correct statement$(s)$:
Statement $I$: There is dimer formation when undergoing freezing.
Statement $II$: There is no change when undergoing boiling.
Statement $III$: Reverse of $I$ and $II$.
Statement $IV$: Dimer formation in freezing and boiling state.

The solution of sugar in water contains

Properties such as boiling point,freezing point,and vapour pressure of a pure solvent change when solute molecules are added to get a homogeneous solution. These are called colligative properties. Applications of colligative properties are very useful in day-to-day life. One of its examples is the use of an ethylene glycol and water mixture as an anti-freezing liquid in the radiator of automobiles.
$A$ solution $M$ is prepared by mixing ethanol and water. The mole fraction of ethanol in the mixture is $0.9$.
Given: Freezing point depression constant of water $(K_{f}^{\text{water}}) = 1.86 \ K \ kg \ mol^{-1}$
Freezing point depression constant of ethanol $(K_{f}^{\text{ethanol}}) = 2.0 \ K \ kg \ mol^{-1}$
Boiling point elevation constant of water $(K_{b}^{\text{water}}) = 0.52 \ K \ kg \ mol^{-1}$
Boiling point elevation constant of ethanol $(K_{b}^{\text{ethanol}}) = 1.2 \ K \ kg \ mol^{-1}$
Standard freezing point of water $= 273 \ K$
Standard freezing point of ethanol $= 155.7 \ K$
Standard boiling point of water $= 373 \ K$
Standard boiling point of ethanol $= 351.5 \ K$
Vapour pressure of pure water $= 32.8 \ mm \ Hg$
Vapour pressure of pure ethanol $= 40 \ mm \ Hg$
Molecular weight of water $= 18 \ g \ mol^{-1}$
Molecular weight of ethanol $= 46 \ g \ mol^{-1}$
In answering the following questions,consider the solutions to be ideal dilute solutions and solutes to be non-volatile and non-dissociative.
$1.$ The freezing point of the solution $M$ is
$(A) \ 268.7 \ K \ (B) \ 268.5 \ K$
$(C) \ 234.2 \ K \ (D) \ 150.9 \ K$
$2.$ The vapour pressure of the solution $M$ is
$(A) \ 39.3 \ mm \ Hg \ (B) \ 36.0 \ mm \ Hg$
$(C) \ 29.5 \ mm \ Hg \ (D) \ 28.8 \ mm \ Hg$
$3.$ Water is added to the solution $M$ such that the mole fraction of water in the solution becomes $0.9$. The boiling point of this solution is
$(A) \ 380.4 \ K \ (B) \ 376.2 \ K$
$(C) \ 375.5 \ K \ (D) \ 354.7 \ K$
Give the answer for questions $1, 2$ and $3.$

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo