Class 12th Chemistry 2017 Set3 Outside Delhi Board Paper Solution
(b) Define the following terms:
(i) Molality (m)
(ii) Abnormal molar mass
OR
(a) 30g of Urea (M = 60 g mol-1) is dissolved in 846 g of water. calculate the vapour pressure of water for this solution if vapour pressure of pure water at 298 K is 23.8 mm Hg.
(b) Write two difference between ideal solution and non-ideal solutions.
(b)
(i) Molality is defined as the number of moles of solute per kilogram of solvent. The SI unit for molality is mol/kg.
(ii) Abnormal molar masses and colligative properties. For the solute which undergo association or dissociation, observed value of colligative property is different from the calculated value of colligative property.
OR
(a) Vapour pressure of water, p1 = 23.8 mm of Hg
Weight of water = 846 g
Weight of urea = 30 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
Number of moles of water n1 = 846 / 18 = 47
Number of mole of urea n2 = 30 / 60 = 0.5
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
(P10 - P1) / P10 = n2 / (n1-n2)
Plug the values we get
(23.8 - p1) / 23.8 = 0.5 / (47+0.5)
(23.8 - p1) / 23.8 = 0.5106
after cross multiply
23.8 – p1 = 23.8 × 0.5106
Solve it we get p1 = 11.6 mm Hg
So, Vapour pressure of water in the given solution = 11.6 mm of Hg
(b)
| Ideal Solution | Non-ideal Solution |
| Obey Raoult's law at every range of concentration. | Do not obey Raoult's law. |
 Neither heat is evolved nor absorbed during dissolution. |
 Endothermic dissolution; heat is absorbed. |
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