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>> Example based on Lowering in Vapour Pressure
Calculate the vapour pressure lowering caused by addition of 50 g of sucrose (molecular mass = 342) to 500 g of water if the vapour pressure of pure water at 25°C is 23.8 mm Hg.
According to Raoult’s law,
p0–ps/p0 = n/n+N
or Λp n/n+N.p0
Given n = 50/342 = 0.416, N = 500/18 = 27.78 and p0 = 23.8
Substituting the values in the above equation,
¦p = 0.146/0.146+27.78 × 23.8 = 0.124 mm Hg
The vapour pressure of pure benzene at a certain temperature is 640mm Hg. A non-volatile solid weighing 2.175 g is added to 39.0 g of benzene. The vapour pressure of the solution is 600 mm Hg. What is the molecular mass of the solid substance?
According to Raoult’s law.
p0–ps/p0 = n/n+N
Let m be the molecular mass of the solid substance.
n = 2.175/m ; N = 39/78 = 0.5
[molecular mass of benzene = 78]
Substituting the values in above equation.
640–600/640 = 2.175/m/2.175/m+0.5 = 2.174/2.175+0.5m
m = 2.175×16 – 2.175/0.5 = 65.25
A solution containing 30 g of a non-volatile solute in exactly 90 g of water has a vapour pressure of 21.85 mm of Hg at 25°C. Further 18 g of water is then added to the solution; the new vapour pressure becomes 22.15 mm Hg of at 25°C. Calculate (a) molecular mass o the solute and (b) vapour pressure of water at 25°C.
Let the vapour pressure of water at 25°C be p0 and molecular mass of the solute be m.
Using Raoult’s law in the following form.
For solution (I), (p0–21.85)/21.85 = 30×18/90×m …(i)
For solution (II), (p0–22.15)/22.15 = 30×18/108×m …(ii)
Dividing Eq. (i) by Eq. (ii),
(p0–21.85)/21.85 × 22.15/(p0–22.15) = 108/90 = 6/5
Substituting the value of p0 in Eq. (i)
M = 67.9
What mass of non-volatile solute (urea) needs to be dissolved in 100 g of water in order to decrease the vapour pressure of water by 5%. What will be the molality of solution?
Using Raoult’s law in the following form,
p0–ps/ps = wM/Wm
If p0 = 100 mm, then ps = 75 mm
100–75/75 = w×18/100×60
w = 111.1
Molality = w×1000/m×W = 111.1×1000/60×100 = 18.52 m