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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