For ethanol-acetone mixture solute-solvent interaction is weaker than solute-solute and solvent-solvent interaction. Give the vapour pressure-mole fraction graph for this solution.

In ethanol-acetone mixture solute-solvent interaction is weaker than solute-solute and solvent-solvent interaction. Therefore, the molecules in the solution have a greater tendency to escape out of the surface. Vapour pressure of the components in the mixture is more than that expected from Raoult's law. Hence, this mixture does not obey Raoult's law, but shows a positive deviation from Raoult's law, when P A > P ∘ A x A and P B > P ∘ B x B , as the total vapour pressure P ∘ A x A + P ∘ B x B is greater than what it should be according to Raoult’s Law. Where, P A , P ∘ A , x A are the vapour pressure, vapour pressure of pure solvent and mole fraction of component A respectively, and that of B be P B , P ∘ B , x B . The vapour pressure-mole fraction graph is as shown below:

For ethanol-acetone mixture solute-solvent interaction is weaker than solute-solute and solvent-solvent interaction. Does this solution obey Raoult's law ?

In ethanol-acetone mixture solute-solvent interaction is weaker than solute-solute and solvent-solvent interaction. Therefore, the molecules in the solution have a greater tendency to escape out of the surface. Vapour pressure of the components in the mixture is more than that expected from Raoult's law. Hence, this mixture does not obey Raoult's law, but shows a positive deviation from Raoult's law.

Complete the table by giving the value of van't Hoff's factor i for complete dissociation of solute. Salt van't Hoff's factor i for complete dissociation of solute. NaCl Al NO 3 3 K 2 SO 4 Al 2 SO 4 3

The van't Hoff's factor i expresses the extent of association or dissociation of solute in the solution. For the dissociation reaction, the van't Hoff's factor i = 1 + α n - 1 . All the given solutes undergo complete dissociation, therefore, α = 1 . n is the number of ions produced after the dissociation of one molecule of solute. Thus, for (i) NaCl → Na + + Cl - , the van't Hoff's factor i = 1 + α n - 1 = 1 + 1 × 2 - 1 = 2 (ii) Al NO 3 3 → Al 3 + + 3 NO 3 - , the van't Hoff's factor i = 1 + α n - 1 = 1 + 1 × 4 - 1 = 4 (iii) K 2 SO 4 → 2 K + + SO 4 2 - , the van't Hoff's factor i = 1 + α n - 1 = 1 + 1 × 3 - 1 = 3 (iv) Al 2 SO 4 3 → 2 Al 3 + + 3 SO 4 2 - , the van't Hoff's factor i = 1 + α n - 1 = 1 + 1 × 5 - 1 = 5 Salt van't Hoff's factor i for complete dissociation of solute. NaCl 2 Al NO 3 3 4 K 2 SO 4 3 Al 2 SO 4 3 5

A solution contains 15 g urea (molar mass = 60 g mol - 1 ) per litre of solution in water has the same osmotic pressure as a solution of glucose (molar mass = 180 g mol - 1 ) in water. Calculate the mass of glucose present in one litre of its solution.

Given, The weight of urea W Urea = 15 g The molar mass of the urea M Urea = 60 g mol - For isotonic solution π Urea = π Glucose Then, C Urea = C Glucose n Urea = n Glucose ⇒ n = w m = 15 60 = 1 4 Assuming the weight of the glucose is = x 1 4 = x 180 x = 45 Therefore, the mass of glucose = 45 g .

What is reverse osmosis? Mention one of its applications.

The process of osmosis can be reversed if a pressure larger than the osmotic pressure is applied to the solution side. As a result, the solvent starts moving from the solution towards the pure solvent through the semipermeable membrane. This process of movement of a solvent through a semipermeable membrane from the solution to the pure solvent by applying excess pressure on the solution side is called reverse osmosis. Reverse osmosis is used for the desalination of seawater to get drinking water from it.

A 5 % solution (by mass) of cane sugar in water has freezing point of 271 K . Calculate the freezing point of 5 % glucose in water if freezing point of pure water is 273 . 15 K .

Sugar cane contains sucrose thus molar mass of sugar = 342 g mol - 1 The 5 % of sugarcane solution means 5 g of sucrose in 100 g of solution. Weight of solvent can be calculated as follows: ∴ w 1 = 100 - 5 = 95 g Molality of sugar solution = w 2 / M 2 w 1 / 1000 = 5 × 1000 342 × 95 = 0 . 1538 m ∆ T f for sugar solution = 273 . 15 - 271 = 2 . 15 K ∆ T f = K f × m K f = 2 . 15 0 . 1538 = 13 . 9791 Km - 1 Molar mass of glucose is 180 g / mol The 5 % of glucose solution means 5 g of glucose in 100 g of solution. Weight of solvent can be calculated as follows. ∴ w 1 = 100 - 5 = 95 g molality = w 2 / M 2 w 1 / 1000 = 5 180 × 1000 95 = 0 . 2924 m ∆ T f = K f × m = 2 . 15 0 . 1538 × 0 . 2924 = 4 . 09 K Freezing point of glucose = 273 . 15 - 4 . 09 = 269 . 06 K

E M B I B E

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR CHEMISTRY>Chapter -1 - Solutions>Kerala Board-2018>Q 2

MEDIUM

12th Kerala Board

IMPORTANT

Earn 100

Define minimum boiling azeotropes with example.

Chapter-Wise 3D Chemistry Videos from Embibe Experts EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR CHEMISTRY

08 min 23 sec

Azeotropic Mixtures

Certain liquid mixtures have a definite composition and boil at a constant temperature. They are called azeotropes. Let’s explore the cause of their behaviour!

04 min 56 sec

Ideal Solutions

The intermolecular forces acting between the components of a solution determine its various characteristics. Let's explore what happens in the case of an ide...

03 min 23 sec

Ideal and Non-ideal Solutions

Any solution that obeys Raoult's law at all temperatures and concentrations, and for the change in enthalpy and the change in volume on mixing is zero is cal...

Learn and Prepare for any exam you want !

|

|

|

|

|

|

|

|