60 gram of urea was dissolved in 9.9 moles of water is the vapour pressure of the pure water is p not the vapour pressure of the solution is
Answers
Answer:
The vapour pressure of the solution is 0.90p.
Explanation:
Given data:
Mass of solute i.e., urea = 60 g
No. of mole of water(solvent), n2 = 9.9 moles
Vapour pressure of pure water, P°solvent = p
To find: Vapour pressure of the solution, Psolution
Let the no. of moles of urea(solute) be “n1”.
Therefore,
Moles of urea, n1 = (mass of urea) / (molar mass of urea) = 60/60.06 = 0.99
Now,
Mole fraction of water(solvent),
χ solvent = n2 / (n1 + n2) = 0.99 / (0.99 + 9.9) = 0.90 …. (i)
According to Raoult’s Law, we have
Psolution = χ solvent * P°solvent
Putting the value of χ solvent from (i) and P°solvent = p, we get
Or, Psolution = 0.90 * p
Answer:
0.90 p0
Explananation Moles of water N= 9.9 moles
Moles of urea n=W/M= 60/60 = 1 Now P0-Ps/P0 = n/N P0-Ps/P0 = 1/9.9 9.9P0- 9.9Ps = P0 8.9P0 - 9.9Ps = P0 8.9 P0 = 9.9 Ps Ps = 8.9/9.9 P0 ≈ 0.90 Ps .