Friday, May 3, 2013

Vapour pressure of pure water at 298 K is 23.8 mm Hg. 50 g of urea NH2CONH2)


Vapour pressure of pure water at 298 K is 23.8 mm Hg. 50 g of urea NH2CONH2) is dissolved in 850 g of water. Calculate the vapour pressure of water for this solution and its relative lowering.
Solution
Given that
Vapour pressure of water,p1 = 23.8 mm of Hg
Weight of water    = 850 g
Weight of urea      = 50 g
Molecular weight of water(H2O) = 1 × 2  + 16  =  18 g mol−1
Molecular weight of urea(NH2CONH2)= 2N + 4H  + C + O
 = 2 × 14 + 4 × 1 + 12 + 16
 = 60 g mol−1
Use formula 
 
Number of moles of water n1  = 850 / 18  = 47.22
Number of mole of urea n2= 50/60 = 0.83
  
Now, we have to calculate vapour pressure of water in the solution. We take vapour
pressure as p1.
Use the formula of  Raoult’s law
Plug the values we get
Cross multiply
23.8 – p1 = 23.8 × 0.0173
Solve it we get
p1 = 23.4 mm Hg
 
Answer
Vapour pressure of water in the given solution = 23.4 mm of Hg
Relative lowering = 0.0173.

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