$A$ substance '$X$' $(1.5 \ g)$ dissolved in $150 \ g$ of a solvent '$Y$' (molar mass $= 300 \ g \ mol^{-1}$) led to an elevation of the boiling point by $0.5 \ K$. The relative lowering in the vapour pressure of the solvent '$Y$' is . . . . . . $\times 10^{-2}$. (Nearest integer)
[Given : $K_{b}$ of the solvent $= 5.0 \ K \ kg \ mol^{-1}$]
Assume the solution to be dilute and no association or dissociation of $X$ takes place in solution.
[Given : $K_{b}$ of the solvent $= 5.0 \ K \ kg \ mol^{-1}$]
Assume the solution to be dilute and no association or dissociation of $X$ takes place in solution.
- A$3$
- B$1$
- C$5$
- D$2$
Explore More
Similar Questions
Assertion : Lowering of vapour pressure is directly proportional to osmotic pressure of the solution.
Reason : Osmotic pressure is a colligative property.
Reason : Osmotic pressure is a colligative property.
$A$ 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 $= 100^{\circ} C$ )
Calculate the freezing point in $^oC$ of an aqueous solution of urea that boils at $100.18\,^oC$. $(K_f = 1.86\, ^oC\, m^{-1}, K_b = 0.512\,^oC\, m^{-1})$
Medium
View SolutionConsider the following aqueous solutions.
$I$. $2.2 \ g$ Glucose in $125 \ mL$ of solution.
$II$. $1.9 \ g$ Calcium chloride in $250 \ mL$ of solution.
$III$. $9.0 \ g$ Urea in $500 \ mL$ of solution.
$IV$. $20.5 \ g$ Aluminium sulphate in $750 \ mL$ of solution.
The correct increasing order of boiling point of these solutions will be:
[Given: Molar mass in $g \ mol^{-1}$: $H=1, C=12, N=14, O=16, Cl=35.5, Ca=40, Al=27, S=32$]
$I$. $2.2 \ g$ Glucose in $125 \ mL$ of solution.
$II$. $1.9 \ g$ Calcium chloride in $250 \ mL$ of solution.
$III$. $9.0 \ g$ Urea in $500 \ mL$ of solution.
$IV$. $20.5 \ g$ Aluminium sulphate in $750 \ mL$ of solution.
The correct increasing order of boiling point of these solutions will be:
[Given: Molar mass in $g \ mol^{-1}$: $H=1, C=12, N=14, O=16, Cl=35.5, Ca=40, Al=27, S=32$]
DifficultJEE MAIN 2026
View SolutionAn aqueous solution freezes at $-0.186 \ ^\circ C$ ($K_f = 1.86 \ ^\circ C \ kg \ mol^{-1}$; $K_b = 0.512 \ ^\circ C \ kg \ mol^{-1}$). What is the elevation in boiling point of the same solution?
Medium
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo