$A$ solution of urea in water has a boiling point of $100.18^{\circ} C$. What is the freezing point of the same solution,if $K_{f}$ and $K_{b}$ of water are $1.86$ and $0.52 \ K \ kg \ mol^{-1}$,respectively (in $^{\circ} C$)? (Boiling point of water $= 100^{\circ} C$ )

  • A
    $-0.34$
  • B
    $-0.22$
  • C
    $-0.64$
  • D
    $-0.32$

Explore More

Similar Questions

An aqueous solution of $NaCl$ shows the depression of freezing point of water equal to $0.372 \, K$. The boiling point of $BaCl_2$ solution of same molality will be .........$^oC$. $[K_f(H_2O) = 1.86 \, K \, kg \, mol^{-1}; K_b(H_2O) = 0.52 \, K \, kg \, mol^{-1}]$

$A$ cylinder containing an ideal gas ($0.1 \; mol$ in $1.0 \; dm^{3}$) is in thermal equilibrium with a large volume of $0.5 \; m$ (molal) aqueous solution of ethylene glycol at its freezing point. If the stoppers $S_{1}$ and $S_{2}$ (as shown in the figure) are suddenly withdrawn,the volume of the gas in litres after equilibrium is achieved will be ............ $litre$.
(Given: $K_{f}$ (water) $= 2.0 \; K \; kg \; mol^{-1}$,$R = 0.08 \; dm^{3} \; atm \; K^{-1} \; mol^{-1}$,freezing point of water $= 273 \; K$)

What type of solution is iodine in air?

Drinking water contains some salt impurities dissolved in it. When this solution is heated in an open container,vapours are formed and separated slowly. In this process,the freezing point and osmotic pressure of the remaining solution will continuously

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