Home
Class 12
CHEMISTRY
Vapour pressure of pure water at 298 K ...

Vapour pressure of pure water at `298 K ` is `23.8 mm Hg`. `50g` of urea `(NH_(2)CONH_(2))` is dissolved in `850g` of water. Calculate the vapour pressure of water for this solution and its relative lowering.

Text Solution

Verified by Experts

Moles of urea `= (50)/(60) = 0.833`
Moles of water `= (850)/(18) = 47.222`
Mole fraction of urea `= (0.833)/(0.833 + 47.222) = 0.0173`
`(p_(A)^(@) - p_(A))/(p_(A)^(@)) = x_(B) = 0.0173`
`(23.8-p_(A))/(23.8) = 0.0173`
`23.8-p_(A) = 0.412`
`p_(A) = 23.39`
Relative lowering of vapour pressure = 0.0173.
Promotional Banner

Topper's Solved these Questions

  • SOLUTIONS

    MODERN PUBLICATION|Exercise NCERT FILE (TEXTBOOK EXERCISES )|39 Videos

  • SOLUTIONS

    MODERN PUBLICATION|Exercise NCERT EXEMPLAR PROBLEMS (MULTIPLE CHOICE QUESTIONS (TYPE -I))|26 Videos

  • SOLUTIONS

    MODERN PUBLICATION|Exercise CONCEPTUAL QUESTIONS|44 Videos

  • SOLID STATE

    MODERN PUBLICATION|Exercise UNIT PRACTICE TEST|13 Videos

  • SURFACE CHEMISTRY

    MODERN PUBLICATION|Exercise COMPETITION FILE (OBJECTIVE TYPE QUESTIONS) (MATCHING LIST TYPE QUESTIONS)|2 Videos

Similar Questions

Explore conceptually related problems

6g of urea is dissolved in 90g of water. The relative lowering of vapour pressure is equal to

3g urea is dissolved in 45g of water. The relative lowering of vapour pressure is :

The vapour pressure of water at 23^(@)C is 19.8 mm of Hg 0.1 mol of glucose is dissolved in 178.2g of water. What is the vapour pressure (in mm Hg) of the resultant solution?

The vapour pressure of pure water at 20^(@)C is 17.5 mm of Hg. A solution of sucrose is prepared by dissoving 68.4 g of sucrese in 1000 g of water. Calculate the vapour pressure of the solution.

The vapour pressure of water at 23^(@)C is 19.8 mm. 0.1 mole glucose is dissolved in 178.2g of water. What is the vapour pressure (in mm) of the resultant solution ?