All questions of Solutions & Colligative Properties for Chemistry Exam

I for AgNO3 = normal mol.wt/observed mol wt= 1+ α∴ α = i −1= (170/92.64 − 1)= 0.835 = 83.5%

According to Raoult's law, for dilute solutions containing non volatile solute, the relative lowering of vapor pressure, Δp/po is equal to the mole fraction of solute, X2.  * The lowering of vapor pressure is proportional to the mole fraction of solute or inversely related to mole fraction of solvent.

Correct Ans Is B not C

Given:
Boiling point of pure CHCl3 (Tb1) = 61.7°C
Kb for CHCl3 = 3.63°C/m

To find:
Boiling point of solution (Tb2) = ?

Solution:
1. Calculate the molality of the solution:
Molality (m) = moles of solute / kg of solvent
Molar mass of CHCl3 = 12 + 1 + 35.5x3 = 119.5 g/mol
Number of moles of CHCl3 = 15000 g / 119.5 g/mol = 125.52 mol
Molality of CHCl3 = 125.52 mol / 15 kg = 8.36 mol/kg
Molar mass of C12H10 = 156.2 g/mol
Number of moles of C12H10 = 616 g / 156.2 g/mol = 3.95 mol
Mass of solvent (CHCl3) = 15 kg
Molality of solution = 3.95 mol / (15 + 0.616) kg = 0.243 mol/kg

2. Calculate the change in boiling point:
ΔTb = Kb x m
ΔTb = 3.63°C/m x 0.243 mol/kg = 0.88149°C

3. Calculate the boiling point of solution:
Tb2 = Tb1 + ΔTb
Tb2 = 61.7°C + 0.88149°C = 62.58°C

Answer:
The boiling point of the solution is 62.58°C, option (D) is correct.

Explanation:

Relative lowering of vapour pressure is given by the formula:

(Relative lowering of vapour pressure) = (Mole fraction of solute) x (Molecular weight of solvent) / (1000 x Density of solvent)

Let the weight of solute be w and weight of solvent A and B be WA and WB respectively.

So, for solvent A:

Mole fraction of solute = w / (MA x WA)
Relative lowering of vapour pressure for A = (w / (MA x WA)) x MA / (1000 x Density of A)

For solvent B:

Mole fraction of solute = w / (MB x WB)
Relative lowering of vapour pressure for B = 2 x (w / (MB x WB)) x MB / (1000 x Density of B)

It is given that both the solvents have equal weight of solute and the relative lowering of vapour pressure for solution B is twice that of solution A.

Therefore,

(w / (MA x WA)) x MA / (1000 x Density of A) = 2 x (w / (MB x WB)) x MB / (1000 x Density of B)

Simplifying this expression, we get:

MA / WB = 2 x MB / WA

Dividing both sides by MA, we get:

1 / (MA x WB) = 2 / (WA x MB)

1 / MA = 2 / MB

Hence, MB = 2 x MA. Therefore, option B is the correct answer.

Raoult's law is applicable when the forces of attractions between like molecules are more or less identical with those between unlike molecules.

Given information:
- Water and Chlorobenzene are immiscible liquids.
- Their mixture boils at 89C under a reduced pressure of 7.7104 Pa.
- The vapour pressure of pure water at 89C is 7104 Pa.

To find: Weight per cent of Chlorobenzene in the distillate.

Solution:
1. Calculation of the total pressure in the system:
The reduced pressure given is the pressure at which the mixture boils. To find the total pressure in the system, we need to add the vapour pressure of pure water at 89C to the reduced pressure.
Total pressure = Vapour pressure of pure water + Reduced pressure
Total pressure = 7104 Pa + 7.7104 Pa
Total pressure = 7111.7104 Pa

2. Calculation of the composition of the vapour:
We assume that the vapour in equilibrium with the liquid mixture is composed of water and chlorobenzene in the same proportion as they exist in the liquid mixture. Let x be the weight per cent of chlorobenzene in the liquid mixture, then (100-x) is the weight per cent of water in the liquid mixture.

The vapour above the liquid mixture will also have the same composition as the liquid mixture.

Now, using Dalton's law of partial pressures, we can write:
Partial pressure of water in the vapour = mole fraction of water in the vapour x Total pressure
Partial pressure of chlorobenzene in the vapour = mole fraction of chlorobenzene in the vapour x Total pressure

We assume that the mole fraction of water in the vapour is equal to the mole fraction of water in the liquid mixture, and similarly for chlorobenzene.

Let's denote the mole fraction of water in the liquid mixture as y. Then, the mole fraction of chlorobenzene in the liquid mixture is (1-y).

Using the above assumptions and Dalton's law, we get:
Partial pressure of water in the vapour = y x 7111.7104 Pa
Partial pressure of chlorobenzene in the vapour = (1-y) x 7111.7104 Pa

3. Calculation of the weight per cent of chlorobenzene in the distillate:
We know that the mixture boils at 89C under the given reduced pressure. This means that the vapour pressure of the liquid mixture at 89C is equal to the reduced pressure.

Using Raoult's law, we can write:
Vapour pressure of the liquid mixture = mole fraction of water in the liquid mixture x Vapour pressure of pure water at 89C
+ mole fraction of chlorobenzene in the liquid mixture x Vapour pressure of chlorobenzene at 89C

Since water and chlorobenzene are immiscible, the mole fraction of each component in the liquid mixture is equal to its weight per cent in the liquid mixture.

At the boiling point, the vapour pressure of the liquid mixture is equal to the total pressure in the system. Therefore, we can write:
Total pressure = y x 7104 Pa + (1-y) x Vapour pressure of chlorobenzene at 89C

Solving for y, we get:
y = (7104 Pa - 7.7104 Pa)/(7104 Pa - Vapour pressure of chlor

Given information:
- One molal solution of a carboxylic acid in benzene
- Boiling point elevation = 1.518 K
- Kb for benzene = 2.53 K kg mol-1

To determine: Degree of association for dimerization of the acid in benzene

Formula:
ΔTb = Kb ∙ molality ∙ degree of association

where,
ΔTb = elevation of boiling point
Kb = ebullioscopic constant for solvent
molality = moles of solute per kg of solvent
degree of association = extent to which the solute associates or dissociates in solution

Solution:
Let's first calculate the molality of the solution.

Molality (m) = 1 mol of solute / 1000 g of solvent

We need to convert the mass of benzene (solvent) to kg.

Mass of benzene = 1000 g
Mass of solute = 1 mol
Molar mass of solute = unknown

Let's assume the molar mass of the solute to be M.

1 mol of solute = M g of solute
Total mass of solution = 1000 g (benzene) + M g (solute)

Molality (m) = 1 mol / (1000 g + M g)

Now, let's use the formula to calculate the degree of association.

ΔTb = Kb ∙ molality ∙ degree of association
1.518 = 2.53 ∙ (1 mol / (1000 g + M g)) ∙ degree of association

Simplifying the equation,

Degree of association = 1.518 ∙ (1000 g + M g) / (2.53 ∙ M)

We can further simplify the equation by assuming the molar mass of the solute to be twice its empirical formula weight (dimerization assumption).

Molar mass of solute = 2 ∙ Empirical formula weight

Degree of association = 1.518 ∙ (1000 g + 2 ∙ EFw g) / (2.53 ∙ 2 ∙ EFw)
Degree of association = (1.518 ∙ 1000) / (5.06 ∙ EFw + 1000)

Let's assume the empirical formula weight of the acid to be x.

Empirical formula weight (EFw) = x
Molar mass of solute = 2 ∙ x

Substituting the values,

Degree of association = (1.518 ∙ 1000) / (5.06 ∙ x + 1000)

We need to find the value of x that gives a degree of association of 80%.

Degree of association = 0.8 = (1.518 ∙ 1000) / (5.06 ∙ x + 1000)

Solving for x,

x = 90 g/mol

Therefore, the empirical formula of the acid is CH2O2 and the degree of association for dimerization of the acid in benzene is 80%.

Explanation:

The vant Hoff factor, denoted by the symbol 'i', is a measure of the extent of dissociation or association of an electrolyte in a solvent. It is defined as the ratio of the moles of particles formed after dissociation or association to the moles of the electrolyte initially present.

When an electrolyte dissociates into ions in a solvent, the vant Hoff factor is greater than one. This is because each formula unit of the electrolyte breaks down into multiple ions.

On the other hand, when an electrolyte associates or combines with solvent molecules, the vant Hoff factor is less than one. This is because the number of particles formed after association is less than the number of formula units initially present.

Example:

Let's consider the example of table salt (sodium chloride, NaCl) in water. When NaCl dissolves in water, it dissociates into Na+ and Cl- ions. Each formula unit of NaCl dissociates into two ions. Therefore, the vant Hoff factor for NaCl in water is 2.

On the other hand, let's consider the example of acetic acid (CH3COOH) in water. Acetic acid is a weak electrolyte and does not completely dissociate into ions in water. Instead, it associates with water molecules to form hydronium ions (H3O+) and acetate ions (CH3COO-). The number of particles formed after association is greater than the number of formula units of acetic acid initially present. Therefore, the vant Hoff factor for acetic acid in water is less than one.

Conclusion:

In summary, the vant Hoff factor for an electrolyte undergoing dissociation is greater than one, while for an electrolyte undergoing association, it is less than one. Therefore, the correct answer is option 'A' - greater than one and less than one.

Colligative property like elevation in boiling point and depression in freezing point is related to the no of moles or particles of substance, if the moles are larger the colligative properties are large
in this case Na3PO4 dissociate to give 4 ions for the same concentration 1M as:-
Na3PO4 -----> 3 Na+ + PO43-
total 4 ions more than that of other options

-A non-volatile substance refers to a substance that does not readily evaporate into a gas under existing conditions. Non-volatile substances exhibit a low vapor pressure and a high boiling point. Sugar and salt are examples of non-volatile solutes.

More volatile component have more its molecules at vapour Phase than in liquid phase..

So mole fraction of A will be higher in vapor phase than in liquid phase.

Given data:
Vapour pressure of pure water (P°) = 50 mm Hg
Change in vapour pressure (ΔP) = 10 mm Hg
Moles of water (n) = 12 mol

To calculate the moles of Na2SO4 required to lower the vapour pressure of water by 10 mm Hg, we can use the formula:

ΔP/P° = X2
where X2 is the mole fraction of solute (Na2SO4) in the solution.

Rearranging the formula, we get:
X2 = ΔP/P°

Substituting the given values, we get:
X2 = 10/50 = 0.2

Now, we know that the mole fraction of solute is given by:
X2 = moles of solute (Na2SO4)/(moles of solute + moles of solvent)

Since we are dissolving Na2SO4 in water, the solvent is water and the moles of solvent are given as 12 mol. Therefore, we can write:

0.2 = moles of Na2SO4/(moles of Na2SO4 + 12)

On solving, we get:
moles of Na2SO4 = 3

Hence, the correct option is (c) 3 mole.

Option (b)

Solution with highest vapour pressure

Vapour pressure is the pressure exerted by the vapours above the surface of the liquid when the liquid and its vapours are in dynamic equilibrium at a particular temperature. The higher the vapour pressure, the more easily the liquid will evaporate.

To determine which solution has the highest vapour pressure, we need to compare the mole fraction of the solute in each solution. The solution with the lowest mole fraction of solute will have the highest vapour pressure.

Mole fraction of solute

The mole fraction of solute in a solution is given by the following formula:

mole fraction of solute = moles of solute / total moles of solution

Let's calculate the mole fraction of solute for each solution:

a) 0.02 M NaCl at 50°C
NaCl is a strong electrolyte and dissociates completely in water. Therefore, 0.02 M NaCl solution contains 0.02 moles of Na+ ions and 0.02 moles of Cl- ions per litre of solution. The total moles of solution is 0.02 + 0.02 = 0.04 moles per litre.
mole fraction of NaCl = 0.02 / 0.04 = 0.5

b) 0.03 M sucrose at 15°C
Sucrose is a non-electrolyte and does not dissociate in water. Therefore, 0.03 M sucrose solution contains 0.03 moles of sucrose per litre of solution. The total moles of solution is 0.03 moles per litre.
mole fraction of sucrose = 0.03 / 0.03 = 1

c) 0.005 M CaCl2 at 50°C
CaCl2 is a strong electrolyte and dissociates completely in water. Therefore, 0.005 M CaCl2 solution contains 0.01 moles of Ca2+ ions and 0.01 moles of Cl- ions per litre of solution. The total moles of solution is 0.01 + 0.01 = 0.02 moles per litre.
mole fraction of CaCl2 = 0.005 / 0.02 = 0.25

d) 0.005 M CaCl2 at 25°C
The mole fraction of CaCl2 in this solution will be the same as in solution c) because the concentration of the solution is the same.

Conclusion

Comparing the mole fractions of solute in each solution, we can see that solution c) 0.005 M CaCl2 at 50°C has the lowest mole fraction of solute (0.25) and therefore the highest vapour pressure. This means that CaCl2 solution will evaporate more easily than the other solutions.

Explanation:

Osmotic pressure is defined as the pressure required to stop the flow of solvent molecules across a semipermeable membrane from a region of lower solute concentration to a region of higher solute concentration.

The osmotic pressure is directly proportional to the concentration of the solute in the solution. Therefore, the higher the concentration of the solute, the higher will be the osmotic pressure.

Among the given options, BaCl2 has the highest osmotic pressure at 25°C. This is because BaCl2 is a strong electrolyte, which dissociates completely into Ba2+ and 2Cl- ions in aqueous solution.

BaCl2 → Ba2+ + 2Cl-

Therefore, the concentration of solute particles in 0.1 M BaCl2 solution is three times higher than that in 0.05 M NaCl or Na2SO4 solutions, and twice as high as that in 0.05 M FeCl3 solution.

Hence, the correct answer is option C, 0.1 M BaCl2.

It is a chemical rxn.When we mix water with acid product wiil be in ionic form. so we cannot separate .

Given data:
Empirical formula of the compound = C10H8Fe
Mass of the compound = 0.26 g
Mass of benzene = 11.2 g
Boiling point of benzene = 80.10C
Boiling point of solution = 80.26C
Kb of benzene = 2.53C/molal

To find: Molecular formula of the compound

Steps involved:

1. Calculate the molality of the solution
2. Calculate the change in boiling point of the solution
3. Calculate the number of moles of the compound in the solution
4. Calculate the molar mass of the compound
5. Find the molecular formula of the compound

1. Calculation of Molality:
Molality of the solution, m = (moles of solute) / (mass of solvent in kg)
Here, mass of solvent = 11.2 g = 0.0112 kg
Empirical formula weight of C10H8Fe = 160 g/mol
Moles of C10H8Fe in 0.26 g = 0.26 g / 160 g/mol = 0.001625 mol
Molality (m) = 0.001625 mol / 0.0112 kg = 0.145 mol/kg

2. Calculation of Change in Boiling Point (∆Tb):
∆Tb = Kb x m
Here, Kb = 2.53C/molal and m = 0.145 mol/kg
∆Tb = 2.53 C/molal x 0.145 mol/kg = 0.366 C

3. Calculation of Moles of the Compound:
∆Tb = Kb x m = (Kb x w2) / (Mf x Kb)
Here, w2 = mass of the compound = 0.26 g
Mf = molar mass of the compound
0.366 C = (2.53 C/molal x 0.26 g) / (Mf x 0.0112 kg)
Mf = 346.3 g/mol
Moles of the compound = 0.26 g / 346.3 g/mol = 0.000750 mol

4. Calculation of Molar Mass of the Compound:
Empirical formula weight of C10H8Fe = 160 g/mol
Molecular formula weight of the compound = 346.3 g/mol
Ratio of molecular formula weight to empirical formula weight = 346.3 g/mol / 160 g/mol = 2.165
The molecular formula of the compound should have a ratio of 2.165 times the empirical formula weight.
Therefore, the molecular formula of the compound = 2.165 x C10H8Fe = C20H16Fe2

Hence, the correct answer is option D, C20H16Fe2.

Explanation:

The Vant Hoff factor is a measure of how many particles a solute dissociates into when it is dissolved in a solvent. For example, NaCl dissociates into two ions (Na+ and Cl-) when it dissolves in water, so its Vant Hoff factor is 2.

In the case of a solute that undergoes dimerisation and trimerisation, the Vant Hoff factor will be affected. Let's consider each case separately:

Dimerisation

When a solute undergoes dimerisation, it means that two molecules of the solute combine to form a dimer. This reduces the number of particles in solution, which in turn affects the Vant Hoff factor.

The minimum value of the Vant Hoff factor for dimerisation is 0.5. This is because the solute has combined into two particles, so the Vant Hoff factor is halved.

Trimerisation

When a solute undergoes trimerisation, it means that three molecules of the solute combine to form a trimer. This further reduces the number of particles in solution, which affects the Vant Hoff factor even more.

The minimum value of the Vant Hoff factor for trimerisation is 0.33. This is because the solute has combined into three particles, so the Vant Hoff factor is reduced to one third.

Overall, the minimum values of the Vant Hoff factors for dimerisation and trimerisation are 0.5 and 0.33, respectively. This means that when a solute undergoes both processes, the minimum value of the Vant Hoff factor will be 0.5 for dimerisation and 0.33 for trimerisation, giving a total minimum Vant Hoff factor of 0.5 + 0.33 = 0.83. However, option D suggests 0.5 and 0.33 as the minimum values for dimerisation and trimerisation, respectively, which is correct.

Answer : 

• d)
  All are correct statements.

Solution:

Given, boiling point elevation = 0.01C
Molal elevation constant, Kf = 0.5 K Kg mol-1
Mass of water, w2 = 100 g
Number of glucose molecules in solution = x x 1021

We know that the boiling point elevation is given by the equation:

ΔTb = Kf × m

where ΔTb is the boiling point elevation, Kf is the molal elevation constant and m is the molality of the solution.

To find the molality of the solution, we need to know the mass of glucose in the solution and the mass of water in the solution.

Let the mass of glucose in the solution be w1.

Then, the total mass of the solution is given by:

w = w1 + w2

The molality of the solution is given by:

m = nglucose / mwater

where nglucose is the number of moles of glucose and mwater is the mass of water in kg.

The number of moles of glucose is given by:

nglucose = w1 / Mglucose

where Mglucose is the molar mass of glucose.

Substituting the values, we get:

ΔTb = Kf × nglucose / mwater

Solving for nglucose, we get:

nglucose = ΔTb × mwater / Kf

Substituting the given values, we get:

nglucose = 0.01 × 100 / 0.5

nglucose = 2

The number of moles of glucose in the solution is 2.

The number of glucose molecules in the solution is given by:

nglucose = N / Avogadro's number

where N is the number of glucose molecules and Avogadro's number is 6.022 x 1023 mol-1.

Substituting the values, we get:

N = nglucose × Avogadro's number

N = 2 × 6.022 x 1023

N = 1.2044 x 1024

The number of glucose molecules in the solution is 1.2044 x 1024.

Rounding off to one significant figure, we get:

x = 1.2

Therefore, the value of x is 1.2.