Question

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%

Answer 1

Answer:

Explanation:

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%?

Report by Hemanuj7athrit 16

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.

Answer 2

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.