Example based on Lowering in the Vapour Pressure

Example:

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.

Solution: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   

Example: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?  

Solution: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

Example: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.  

Solution: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

Example: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?  

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

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