At a certain temperature,$5 \ g$ of a non-electrolyte solute is dissolved in $100 \ g$ of water. The vapour pressure of the resulting solution is $2985 \ N/m^2$,and the vapour pressure of pure water is $3000 \ N/m^2$. Calculate the molar mass of the solute.
- A$60$
- B$120$
- C$180$
- D$380$
Explore More
Similar Questions
Relative lowering in vapour pressure of a solution containing a non-volatile solute is the ratio of:
Explain how the solubility of a gaseous solute and the vapour pressure of a liquid solution follow Henry's and Raoult's laws.
Difficult
View Solution$1 \ mol$ of heptane $(V.P = 92 \ mm \ Hg)$ is mixed with $4 \ mol$ of octane $(V.P = 31 \ mm \ Hg)$. The vapor pressure of the resulting ideal solution is .......... $mm \ Hg$.
Medium
View SolutionThe vapour pressure of pure liquid solvent is $0.50 \ atm$. When a non-volatile solute $B$ is added to the solvent,its vapour pressure drops to $0.30 \ atm$. The mole fraction of the solute $B$ is $......$
Easy
View SolutionThe vapour pressure of acetone $(CH_3COCH_3)$ at $20\,^{\circ}C$ is $185\, torr$. When $1.2\, g$ of a non-volatile substance was dissolved in $100\, g$ of acetone at $20\,^{\circ}C$,its vapour pressure was $183\, torr$. The molar mass $(g\, mol^{-1})$ of the substance is
Easy
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