Vapour
pressure of pure water at 298 K is 23.8 mm Hg. 50 g of urea NH2CONH2)
is dissolved in 850 g of water. Calculate the vapour pressure of water for this
solution and its relative lowering.
Solution
Given that
Vapour pressure of water,p1
= 23.8 mm of Hg
Weight of water = 850 g
Weight of urea = 50 g
Molecular weight of water(H2O)
= 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 of water n1 = 850 / 18
= 47.22
Number of mole of urea n2
= 50/60 = 0.83
Now, we have to calculate vapour pressure
of water in the solution. We take vapour
pressure as p1.
Use the formula of Raoult’s law
Plug the values we get
Cross multiply
23.8 – p1 = 23.8 × 0.0173
Solve it we get
p1 = 23.4 mm Hg
Answer
Vapour pressure of water in the
given solution = 23.4 mm of Hg
Relative lowering =
0.0173.
Nice explanation!
ReplyDeleteCan i ignore 0.83 in denominator.
ReplyDeleteNope
DeleteKar sakte hai it is in very less amount
DeleteYes you can bcoz it very less than n1 soo it can be taken negligible
DeleteI don't know why you make lengthy solutions, it can be made easily in 5 steps only
Delete@Admin
ReplyDeletePlease explain more about Vapour pressure of pure water
Can anyone can answer this question?6g of urea was dissolved in9.9moles of water.If the vapour pressure of water is p0 the vapour pressure of solution is..
ReplyDeleteBut the ans. Is 289.5
ReplyDeleteBut the ans. Is 289.5
ReplyDeleteit is given in bar conver it by mulriplying it bar from mm of hg
DeleteWhy don't we use the formula ∆P = P°x2
ReplyDeleteYou calculated the mole fraction wrong
ReplyDeleteIt should be 0.39
Correct it!