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Vapour pressure of a solution containing 6gm of non volatile solute in 180gm of water is 20torr if 1mol of water is further added its vapour pressure is increased by 0.02torr then calculate vapour pressure of pure water and molar mass of non volatile solute

Vapour pressure of a solution containing 6gm of non volatile solute in 180gm of water is 20torr if 1mol of water is further added its vapour pressure is increased by 0.02torr then calculate vapour pressure of pure water and molar mass of non volatile solute

Grade:12th pass

1 Answers

Arun
25750 Points
3 years ago
vapour pressure of a substance above a solution is proportional to the partial fraction (of total moles) of the substance in the solution.    180 gm = 10 moles of water.  Let M be molar mass of solute.    Let k be proportionality constant  in  Torr.
 
10/[10+6/M]  = k 20                      =>   10 = k 20  [10+6/M]
 
11/[11+6/M] = k 20.02                  =>   11 = k 20.02 [11+6/M]
 
     200.2*11 + 200.2*6/M   =   2200 + 220*6/M
 
     2.2 = 19.8*6/M                 => M = 54
 
     4.5 / 91 = k  =>            1/k = 910/45 = 20.2222
 
for pure water the vapoour pressure will be 20.22 torr...
 
Another way,   when mole fraction of water is increased by 0.001 from 0.989 to 0.990 above, the vapor pressure increased by 0.02 torr.    If mole fraction is increased by 0.01 to make that pure water,  the pressure increases by  0.02*10 = 0.2 torr.  So pressure will be 20.2 torr
 
 

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