Question

The vapour pressure of a solution at 298 K
containing 50 g of urea in 850 g of water will
be (Given Pio = 23.8 mm at 298 K)
(NCERT Pg. 49
(1) 22.4 mm
(2) 22.9 mm
(3) 23.1 mm
(4) 23.4 mm​

Answer 1

Answer:

Vapour pressure of water,P1=23.8 m of hg

The weight of water=850 g

     

 The weight of Urea=50 g 

The molecular weight of water(H_2O)=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=molar massmass

number of moles of water   n1=18850=47.22

number of moles of urea   n2=6050=0.83

Now, we have to calculate the vapour pressure of water in the solution. we take vapour pressure as P1.

Use the formula of Raoult's law

      P10P10−P1=n1+n2n2

plug 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