At room temperature,a dilute solution of urea is prepared by dissolving $0.60 \ g$ of urea in $360 \ g$ of water. If the vapour pressure of pure water at this temperature is $35 \ mm \ Hg$,the lowering of vapour pressure will be: .............. $mm \ Hg$ (molar mass of urea $= 60 \ g \ mol^{-1}$)

  • A
    $0.027$
  • B
    $0.031$
  • C
    $0.028$
  • D
    $0.017$

Explore More

Similar Questions

Calculate the relative lowering of vapour pressure of a solution containing $46 \ g$ of non-volatile solute in $162 \ g$ of water at $20^{\circ} C$. [Molar mass of non-volatile solute $= 46 \ g \ mol^{-1}$]

The boiling points of $C_6H_6$,$CH_3OH$,$C_6H_5NH_2$,and $C_6H_5NO_2$ are $80 \, ^\circ C$,$65 \, ^\circ C$,$184 \, ^\circ C$,and $212 \, ^\circ C$ respectively. Which one exhibits the maximum vapor pressure at room temperature?

What is the mole ratio of benzene $(P_B^o = 150\, torr)$ and toluene $(P_T^o = 50\, torr)$ in the vapour phase if the given solution has a vapour pressure of $120\, torr$?

Two liquids $A$ and $B$ have $P_A^o$ and $P_B^o$ in the ratio of $1 : 3$ and the ratio of number of moles of $A$ and $B$ in the liquid phase is $1 : 3$. The mole fraction of $A$ in the vapour phase in equilibrium with the solution is equal to:

Dry air was drawn through bulbs containing pure water,then through bulbs containing a solution of $600 \ g$ of a non-electrolyte in $360 \ g$ of water at the same temperature,and finally through a tube in which dried $CaCl_2$ was placed. The solution bulb gained $1.5 \ g$ and the dried $CaCl_2$ gained $2 \ g$. The molecular mass of the solute is:

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