Home
Class 12
CHEMISTRY
Relative decrease in vapour pressure in ...

Relative decrease in vapour pressure in vapour pressure of an aqueous solution NaCI is 0.1067, Number of moles of NaCI present in 180 g `H_(2)O` is:

A

2 mol

B

1 mol

C

3 mol

D

4 mol

Text Solution

Verified by Experts

The correct Answer is:
B
`(Deltap)/(p) = x_(B)`
`x_(B) = 0.167`
No of moles of water `= (180)/(18) = 10` mol
Let no of moles of `NaCI = n`
Van't Hoff factor i for `NaCI = 2`
`:. x_(B) = (2n)/(2n+10)`
`(2n)/(2n+10) = 0.167`
`(n)/(n+5) = 0.167`
`n = 0.167 n + 0.845`
`0.833 n = 0.835`
`n = (0.835)/(0.833) ~~ 1`
Promotional Banner

Topper's Solved these Questions

  • SOLUTIONS

    DINESH PUBLICATION|Exercise REVISION QUESTION|186 Videos

  • SOLUTIONS

    DINESH PUBLICATION|Exercise OBJECTIVE TYPE MCQs|47 Videos

  • SOLID STATE

    DINESH PUBLICATION|Exercise Brain storming|10 Videos

  • SOME BASIC CONCEPTS OF CHEMISTRY

    DINESH PUBLICATION|Exercise ULTIMATE PREPARATORY PACKAGE|20 Videos

Similar Questions

Explore conceptually related problems

Relative decrease in vapour pressure of an aqueous NaCl is 0.167 . Number of moles of NaCl present in 180g of H_(2)O is:

Relative lowering in vapour pressure of an aqueous NaCl solution is 1/6. What is the number of moles of NaCl present in 180g H_(2)O

Calculate relative lowering of vapour pressure of 0.161 molal aqueous solution

The relative decreases in the vapour pressure of an aqueous solution containing 2 mol [Cu(NH_(3)_(3)Cl] in 3 mol H_(2)O is 0.50 . On reaction with AgNO_(3) , this solution will form

Relative loweringof vapour pressure of a dilute solution is 0.2 . What is the mole fraction of non-volatile solute ?

If relative decrease in vapour pressure is 0.4 for a solution containing 1 mol NaCl in 3 mol of H_(2)O , then % ionization NaCl is