Calculate the molality of a solution of a non-volatile solute having a depression in freezing point of $0.93 \ K$ and a cryoscopic constant of the solvent of $1.86 \ K \ kg \ mol^{-1}$.
- A$0.3 \ mol \ kg^{-1}$
- B$0.4 \ mol \ kg^{-1}$
- C$0.5 \ mol \ kg^{-1}$
- D$0.6 \ mol \ kg^{-1}$
Explore More
Similar Questions
What is the cryoscopic constant of water if $5 \ g$ of glucose in $100 \ g$ of water causes a depression in freezing point of $2.15 \ K$? (Molar mass of glucose $= 180 \ g \ mol^{-1}$)
Which of the following will produce the maximum depression in freezing point of its aqueous solution?
Easy
View SolutionThe freezing point of an aqueous solution containing $17 \ g$ of $C_2H_5OH$ in $1000 \ g$ of water is ......... $^oC$. (Given $K_f$ of water $= 1.86 \ K \ kg \ mol^{-1}$)
Medium
View SolutionIf the freezing point of an aqueous urea solution is $271.14 \ K$ at $1 \ \text{atm}$ pressure (given $K_f$ of water = $1.86 \ K \ kg/mol$),then what is the mole fraction of urea in this solution? (Freezing point of pure water is $273 \ K$)
Medium
View SolutionIf the freezing point of a $5\%$ (by mass) aqueous solution of cane sugar is $271 \ K$ and the freezing point of pure water is $273.15 \ K$,then the freezing point of a $5\%$ (by mass) aqueous solution of glucose will be .......... $K$.
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