Vapour pressure of water is 360 mm Hg, how much urea should be added to 200 ml water
to reduce its vapour pressure by 0.5%? (Molecular wt. of urea = 60)
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Answer:
Explanation:
Vapor Pressure of pure water = p0=360mm Hg
Lowered Vapor Pressure, p= 360-0.5% of 360
= 360-1.8
= 358.2 mm Hg
Weight of water w1= 200mlx1gml=200g
Weight of Urea w2= ?
Molecular weight of water, M1= 18g/mol
Molecular weight of Urea, M2= 60g/mol
Accoding Roult's Law:
P0-p/p = n2/n1+n2
360-358.2/358.2= w2/M2 divided by w1/m1 + w2/m2
0.005=w2/m2 divided by w1/m1+w2/m2
0.005/0.995 x w1/m1 x M2 = w2
0.005 x 200/18 x 60 = w2
W2 = 3.33g
3.33g Urea should be added.
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