Question

At 298k the vapour pressure of water is 23.75 mm hg. Cal the vapour pressure at same temperature over 5% aqueous solution of urea

Answer 1

Let the mass of solution be 100g

Therefore mass of urea be 5 g

Molecular mass of urea be 60g

Xurea = 5/60= 1/12

Mole fraction of urea

Xurea = (1/2)/((1/2)+5.27)=((1/2)/1+(12x5.27)/12)=1/64.24

Using formula

23.75-p/23.75 = 1/64.24

Or 23.75-p = 23.75/64.24

Or p = 23.75-23.64.24

= 23.75(63.24/64.24)

= 23.75x0.98443

= 23.38 mm Hg

Answer 2

Explanation:

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It is given that vapour pressure of water, P°1 = 23.8 mm of Hg ..

 Weight of water taken, w1 = 850 g.

  Weight of urea taken, w2 = 50g.

Molecular weight of water, M1 = 18 g mol−1  Molecular weight of urea, M2 = 60 g mo^l−1.

Now, we have to calculate vapour pressure of water in the solution. We take vapour pressure as p1.

 Now, from Raoult’s law, we have:

 Use the formula of Raoult's law

      P1°-P1/P1°=n2/n1+n2

put the values we get

23.823.8−P1=47.22+0.830.83

23.823.8−P1=0.0173

23.8−P1=23.8×0.0173

P1=23.4 m Hg..

Vapour Pressure of water in the given solution=23.4 mm of Hg..

Relative lowering=0.0173