chapter 2 -Solutions.

30 Dec 2020 7:44 am

Balbharati solutions, for, Chemistry, 12th, Standard, HSC, Maharashtra, State, Board, chapter 2, Solutions,

Exercise | Q 1.01 | Page 44

Choose the most correct option.

The vapour pressure of a solution containing 2 moles of a solute in 2 moles of water (vapour pressure of pure water = 24 mm Hg) is ______.

24 mm Hg

32 mm Hg

48 mm Hg

12 mm Hg

Solution

12 mm Hg

The colligative property of a solution is _______

vapour pressure

boiling point

osmotic pressure

freezing point

Solution

osmotic pressure.

In calculating osmotic pressure the concentration of solute is expressed in _______.

molarity

molality

mole fraction

mass percent

Solution

molarity

Ebullioscopic constant is the boiling point elevation when the concentration of a solution is _______.

1 m

1 M

1 mass %

1-mole fraction of solute.

Solution

1m

Cryoscopic constant depends on _______.

nature of solvent

nature of solute

nature of solution

number of solvent molecules

Solution

number of solvent molecules.

Identify the CORRECT statement.

Vapour pressure of solution is higher than that of pure solvent.

Boiling point of solvent is lower than that of solution.

Osmotic pressure of solution is lower than that of solvent.

Osmosis is a colligative property.

Solution

Boiling point of solvent is lower than that of solution.

A living cell contains a solution which is isotonic with 0.3 M sugar solution. What osmotic pressure develops when the cell is placed in 0.1 M KCl solution at body temperature?

5.08 atm

2.54 atm

4.92 atm

2.46 atm

solution

2.54 atm.

The osmotic pressure of blood is 7.65 atm at 310 K. An aqueous solution of glucose isotonic with blood has the percentage (by volume)________.

5.41 %

3.54 %

4.53 %

53.4 %

solution

5.41 %

Vapour pressure of a solution is _______.

directly proportional to the mole fraction of the solute

inversely proportional to the mole fraction of the solute

inversely proportional to the mole fraction of the solvent

directly proportional to the mole fraction of the solvent

solution

inversely proportional to the mole fraction of the solute.

Pressure cooker reduces cooking time for food because _______.

boiling point of water involved in cooking is increased

heat is more evenly distributed in the cooking space

the higher pressure inside the cooker crushes the food material 

cooking involves chemical changes helped by a rise temperature

solution

boiling point of water involved in cooking is increased

Henry’s law constant for a gas CH3Br is 0.159 mol dm-3 atm-1 at 25 °C. What is the solubility of CH3Br in water at 25 °C and partial pressure of 0.164 atm?

0.0159 mol L-1

0.164 mol L-1

0.026 M

0.042 M

solution

0.026 M

Which of the following statement is NOT correct for 0.1 M urea solution and 0.05 M sucrose solution?

Osmotic pressure exhibited by urea solution is higher than that exhibited by sucrose solution

Urea solution is hypertonic to sucrose solution

They are isotonic solutions

Sucrose solution is hypotonic to urea solution

solution

They are isotonic solutions.

Exercise | Q 2.01 | Page 45

Answer the following in one or two sentences.

What is osmotic pressure?

solution

The hydrostatic pressure (on the side of solution) that stops osmosis is called an osmotic pressure of the solution.

OR

The excess of pressure on the side of the solution that stops the net flow of solvent into the solution through a semipermeable membrane is called osmotic pressure.

A solution concentration is expressed in molarity and not in molality while considering osmotic pressure. Why?

solution

1. The osmotic pressure measurements are made at a specific constant temperature. Molarity remains constant at a specific temperature.

2. It is not necessary to express concentration in a temperature-independent unit like molality.

Hence, the solute concentration is expressed in molarity while calculating osmotic pressure rather than molality

Write the equation relating boiling point elevation to the concentration of the solution.

solution

The boiling point elevation is directly proportional to the molality of the solution. Thus,

Δ Tb ∝ m

∴ Δ Tb ∝ Kb m

where, m is the molality of solution. The proportionality constant Kb is called boiling point elevation constant or molal elevation constant or ebullioscopic constant.

A 0.1 m solution of K2SO4 in water has a freezing point of – 4.3 °C. What is the value of van’t Hoff factor if Kf for water is 1.86 K kg mol–1?

solution

Given: Molality of K2SO4 solution = m = 0.1 m 

Freezing point of solution = Tf = – 4.3 °C 

Kf of water = 1.86 K kg mol–1

To find: van’t Hoff factor

Formula: ΔTf = i Kf m

Calculation: 

ΔTf = 

Tf0 – Tf

= 0 °C – (- 4.3 °C) = 4.3 °C = 4.3 K

Now, using formula,

ΔTf = i Kf m

What is van’t Hoff factor?

solution

van’t Hoff factor (i) is defined as the ratio of colligative property of a solution of electrolyte divided by the colligative property of nonelectrolyte solution of the same concentration.

How is van’t Hoff factor related to degree of ionization?

Solution

The van’t Hoff factor is related to degree of ionization as follows:

i = 1 + α (n – 1)

or

α = 

i=  i — 1

     n– 1

where, α = Degree of ionization/dissociation

i = van’t Hoff factor

n = Moles of ions obtained from ionization of 1 mole of electrolyte.

Which of the following solution will have higher freezing point depression and why?

i. 0.1 m NaCl

ii. 0.05 m Al2(SO4)3

Solution

NaCl   →   Na+     +    Cl-

0.1 m       0.1 m         0.1 m 

Total particles in solution = 0.2 mol

For 0.05 m Al2(SO4)3:

Al2(SO4)3    →  2Al3+     + 3

SO42-

0.05 m             0.1 m            0.15 m

Total particles in solution = 0.25 mol

Al2(SO4)3 solution contains more number of particles than NaCl solution. Hence, Al2(SO4)3 solution has maximum ΔTf.

Therefore, the freezing point depression of 0.05 m Al2(SO4)3 solution will be higher than 0.1 m NaCl solution.

State Raoult’s law for a solution containing a nonvolatile solute.

Solution

The Raoult’s law states that, “the vapour pressure of solvent over the solution is equal to the vapour pressure of pure solvent multiplied by its mole fraction in the solution.”

What is the effect on the boiling point of water if 1 mole of methyl alcohol is added to 1 dm3 of water? Why?

Solution

i. When 1 mole of methyl alcohol is added to 1 dm3 of water, the boiling point of water decreases.

ii. Methyl alcohol is a volatile liquid. Therefore, it increases the vapour pressure of a solution at a given temperature. Hence, the solution boils at lower temperature.

Which of the four colligative properties is most often used for molecular mass determination? Why?

Solution

i. Among the four colligative properties, osmotic pressure is most often used for molecular mass determination.

ii. Osmotic pressure is much larger and therefore more precisely measurable property than other colligative properties.

Therefore, it is useful to determine molar masses of very expensive substances and of the substances that can be prepared in small quantities.

How vapour pressure lowering is related to a rise in the boiling point of solution?

Solution

i. At the boiling point of a liquid, its vapour pressure is equal to 1 atm.

ii. In order to reach boiling point, the solution and solvent must be heated to a temperature at which their respective vapour pressures attain 1 atm.

iii. At any given temperature the vapour pressure of a solution is lower than that of pure solvent. Hence, the vapour pressure of solution needs a higher temperature to reach 1 atm than that of needed for vapour pressure of solvent.

Therefore, vapour pressure lowering causes a rise in the boiling point of a solution.

What are isotonic and hypertonic solutions?

Solution

i. Isotonic solutions:

Two or more solutions having the same osmotic pressure are said to be isotonic solutions.

e.g. For example, 0.1 M urea solution and 0.1 M sucrose solution are isotonic because their osmotic pressures are equal. Such solutions have the same molar concentrations but different concentrations in g/L. If these solutions are separated by a semipermeable membrane, there is no flow of solvent in either direction.

ii. Hypertonic solution:

If two solutions have unequal osmotic pressures, the more concentrated solution with higher osmotic pressure is said to be the hypertonic solution.

e.g. For example, if osmotic pressure of sucrose solution is higher than that of urea solution, the sucrose solution is hypertonic to urea solution.

Exercise | Q 3.3 | Page 46

Answer the following.

A solvent and its solution containing a nonvolatile solute are separated by a semipermeable membrane. Does the flow of solvent occur in both directions? Comment giving a reason.

Solution

1. When a solution and pure solvent or two solutions of different concentrations are separated by a semipermeable membrane, the solvent molecules pass through the membrane.

2. The passage of solvent molecules through the semipermeable membrane takes place in both directions, since the solvent is on both sides of the membrane.

3. However, the rate of passage of solvent molecules into the solution or from a more dilute solution to more concentrated solution is found to be greater than the rate in the reverse direction.

4. This is favorable since the vapour pressure of solvent is greater than that of solution.

The osmotic pressure of CaCl2 and urea solutions of the same concentration at the same temperature are respectively 0.605 atm and 0.245 atm, calculate van’t Hoff factor for CaCl2.

Solution

Given: Osmotic pressure of CaCl2 solution = 0.605 atm

Osmotic pressure of urea solution = 0.245 atm

To find: The value of van’t Hoff factor

Formulae: π = MRT, π = iMRT

Calculation: For urea solution

π = MRT

0.245 atm = MRT       ….(i)

For CaCl2 solution

π = iMRT

0.602 atm = iMRT       ….(ii)

Explain reverse osmosis.

Solution

i. If a pressure larger than the osmotic pressure is applied to the solution side, then pure solvent from the solution passes into pure solvent side through the semipermeable membrane. This phenomenon is called reverse osmosis.

ii. For example, consider fresh water salt water separated by a semipermeable membrane. When the pressure larger than the osmotic pressure of a solution is applied to solution, pure water from salty water passes into fresh pure water through the membrane. Thus, the direction of osmosis can be reversed by applying a pressure larger than the osmotic pressure.

iii. The schematic set up for reverse osmosis is as follows:

How molar mass of a solute is determined by osmotic pressure measurement?

Solution


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