Colligative Properties

The correct answer is 45
W_solute = 2.5 × 10^–3 kg
W_solvent = 75 × 10^–3 kg
ΔT_b = 373.535 – 373.15= 0.385 K
K_b(H₂O) = 0.52 K kg mol^–1


= 45.02 ≈ 45

Calculation of Depression in Freezing Point:

1. Calculate the moles of AgNO₃:
   Moles of AgNO₃ = Molarity × Volume (L) = 1 × 0.020 = 0.020 mol.
2. Determine the moles of KCl:
   From the reaction:
   AgNO₃ + KCl → AgCl + KNO₃,
   1 mole of AgNO₃ reacts with 1 mole of KCl. Therefore, the moles of KCl are:
   Moles of KCl = 0.020 mol.
3. Calculate the molality of the KCl solution:
   The volume of the solution is 25 mL, and the density is 1 g/mL, so the mass of the solution is:
   Mass of solution = 25 g.
   Since the KCl solution is dilute, the mass of the solvent is approximately:
   Mass of solvent = 25 g = 0.025 kg.
   The molality is:
   Molality = Moles of solute / Mass of solvent (kg) = 0.020 / 0.025 = 0.8 mol/kg.
4. Calculate the depression in freezing point:
   KCl dissociates completely into K⁺ and Cl⁻, so the van ’t Hoff factor (i) is 2.
   The depression in freezing point is given by:
   ∆T_f = i × K_f × Molality
   Substituting the values:
   ∆T_f = 2 × 2.0 × 0.8 = 3.2 K.
   The nearest integer is:
   ∆T_f = 3 K.

Conclusion: The depression in freezing point is 3 K.

For isotonic solution: Now, for : Percentage dissociation: Thus, the percentage dissociation of K_2SO_4 is 75%.

To solve this problem, we need to understand the concept of osmosis and how it affects the molarity of solutions separated by a semipermeable membrane (SPM).

In the given scenario, we have the following setup:

In this arrangement, is on Side X and is on Side Y. Osmosis refers to the movement of solvent molecules through a semipermeable membrane from a region of lower solute concentration to a region of higher solute concentration.

Now, let's analyze the options:

1. Green colour formation observed on side Y.
2. Green colour formation observed on side X.
3. Molarity of solution is lowered.
4. Molarity of solution is lowered.

Since the solutions are separated by a semipermeable membrane, only the solvent (typically water) can move to equalize solute concentrations, not the solute itself. Therefore, any change in molarity due to osmosis would typically be a result of water moving from the side with lower solute concentration to the side with higher solute concentration, making the latter solution more dilute.

In this situation, osmosis will cause the water to move from the side (Side X) to the side (Side Y) because the solution is involved in forming the green complex , thereby reducing its effective concentration.

Thus, the molarity of the solution is lowered as water moves into Side Y to equalize concentration due to osmosis.

Therefore, the correct option is:

Molarity of solution is lowered.

The question focuses on the effect of adding a small quantity of naphthalene to benzene on its freezing point. This concept is related to colligative properties in chemistry, specifically freezing point depression. Here’s a detailed explanation:

Understanding Freezing Point Depression

The freezing point of a pure substance is the temperature at which it changes from a liquid to a solid. When a solute is dissolved in a solvent, the freezing point of the solvent decreases. This phenomenon is known as freezing point depression, a colligative property which depends on the number of solute particles in a solvent, not the nature of the solute itself.

Key Formula

The depression in freezing point can be calculated using the formula:

• is the change in freezing point.
• is the van 't Hoff factor (number of particles the solute splits into).
• is the cryoscopic constant of the solvent (benzene in this case).
• is the molality of the solution.

Application to the Given Problem

When naphthalene is added to benzene:

• Naphthalene acts as a solute and benzene as a solvent.
• The addition of naphthalene particles to benzene results in freezing point depression according to the formula above.

Conclusion

Based on the concept of freezing point depression, when a small quantity of naphthalene is added to benzene, the freezing point of benzene decreases.

Correct Answer: Decreases

For NaCl dissociation:

Concentration of NaCl = 0.05 M.
Effective concentration ( ) = 0.05 M + 0.05 M = 0.1 M.
Glucose concentration ( ) = 0.2 M.
The osmotic pressure:


Substituting:

Nearest integer = 25 bar.

The problem asks to determine the order of the van't Hoff factor (i) for three aqueous solutions of NaCl with different concentrations: 0.1 M (A), 0.01 M (B), and 0.001 M (C).

Concept Used:

The van't Hoff factor ( ) is a measure of the extent of dissociation or association of a solute in a solution. For an electrolyte, it is defined as:

For an electrolyte that dissociates, the van't Hoff factor is related to the degree of dissociation ( ) by the formula:

where is the number of ions produced from one formula unit of the solute.

For strong electrolytes like NaCl, while dissociation is considered almost complete, it is not 100% in practice, especially at higher concentrations. This is due to inter-ionic attractions between the cations and anions in the solution. At higher concentrations, ions are closer together, leading to stronger attractions and the formation of "ion pairs," which reduces the effective number of free-moving particles. As the solution is diluted (concentration decreases), the average distance between ions increases, reducing inter-ionic forces. This leads to a higher degree of dissociation ( ). Consequently, the van't Hoff factor ( ) increases with dilution and approaches the theoretical value ( ) at infinite dilution.

Step-by-Step Solution:

Step 1: Analyze the dissociation of NaCl.

Sodium chloride (NaCl) is a strong electrolyte that dissociates in water into one sodium ion ( ) and one chloride ion ( ).

From one formula unit of NaCl, two ions are produced. Therefore, the theoretical maximum value of the van't Hoff factor is .

Step 2: Relate van't Hoff factor to the degree of dissociation for NaCl.

Using the formula with , we get:

This equation shows that the van't Hoff factor ( ) for NaCl is directly proportional to its degree of dissociation ( ).

Step 3: Analyze the effect of concentration on the degree of dissociation.

The three solutions have the following concentrations:

• Solution A: 0.1 M
• Solution B: 0.01 M
• Solution C: 0.001 M

The order of concentrations is: .

As the concentration of the electrolyte solution decreases, the solution becomes more dilute. In a more dilute solution, the ions are farther apart, and the inter-ionic forces of attraction are weaker. This leads to a greater degree of dissociation ( ).

Therefore, the order of the degree of dissociation for the three solutions will be the reverse of the order of their concentrations:

Step 4: Determine the order of the van't Hoff factor.

Since , the van't Hoff factor ( ) will follow the same order as the degree of dissociation ( ).

Let , , and be the van't Hoff factors for solutions A, B, and C, respectively. Because , it follows that:

Final Result:

The most dilute solution (C) will have the highest degree of dissociation and thus the largest van't Hoff factor, while the most concentrated solution (A) will have the lowest degree of dissociation and the smallest van't Hoff factor. The value of for all three solutions will be less than the theoretical value of 2, but will approach 2 as the concentration decreases.

The order of the van't Hoff factor for these solutions will be .

To find the van’t Hoff factor ( ) for the compound which is 30% ionised, we need to understand how ionisation impacts .

Assume initially we have 1 mole of . Upon ionisation, it dissociates as follows:

This means, for complete ionisation, from 1 mole of , we would get 2 moles of and 1 mole of , equating to 3 moles of ions in total.

However, it is only 30% ionised. Therefore:

• Moles of ionised = 0.3 moles
• Moles of undissociated = 0.7 moles

From 0.3 moles of ionised, ions produced are:

• Moles of = 2 × 0.3 = 0.6 moles
• Moles of = 0.3 moles

Total moles after ionisation = moles undissociated + moles of ions = 0.7 + 0.6 + 0.3 = 1.6 moles

The van’t Hoff factor ( ) is defined as:

To solve this problem, we will use the concept of boiling point elevation. The formula for boiling point elevation is given by:

where is the elevation in boiling point, is the van 't Hoff factor (ionization number), is the ebullioscopic constant, and is the molality of the solution.

Step 1: Calculate Molality for Each Solution

For AX₂:

Dissolved mass

Molar mass

Molality ( ) is calculated as:

For AY₂:

Dissolved mass

Molar mass

Molality ( ) is:

Step 2: Determine the van 't Hoff Factor (i)

Given .

For AX₂:

Boiling point change

The value indicates no ionization, meaning AX₂ is unionized.

For AY₂:

Boiling point change

This suggests AY₂ is ionized, breaking into AY and 2Y⁻ (fully ionized).

Conclusion

Given calculations indicate that AX₂ is completely unionized while AY₂ is fully ionized, which corresponds to the statement: "AX₂ is completely unionised while AY₂ is fully ionised."

To find the dissociation constant for the weak acid HA, we need to use the principle of freezing point depression. Here are the steps for the calculation:

1. The freezing point depression ( ) is given as .
2. We can use the formula for freezing point depression: , where
   • is the van 't Hoff factor.
   • (cryoscopic constant of water).
   • (molality, given as 0.1 m).
3. For a monobasic weak acid dissociating as , the van 't Hoff factor is given by: , where is the degree of dissociation.
4. Substitute the known values into the depression formula:
5. Solving the equation, we get:
6. This simplifies to: , hence .
7. The dissociation constant is related to by: , where .
8. Substituting the values, we get:
9. This yields:
10. Simplifying:
11. Thus, the dissociation constant for the acid is .

Therefore, the correct answer is .

Step 1: Understanding the Dissociation of Salt MX
The salt MX dissociates in water as follows: For each mole of MX , it dissociates to give one mole of M and three moles of X .
Step 2: van't Hoff Factor (i)
The van't Hoff factor is given as . The van't Hoff factor represents the total number of particles in solution per formula unit of solute. In this case, for MX , the dissociation would produce 4 particles (1 M and 3 X ) per formula unit of MX .
However, since , this suggests that the dissociation is not complete, and the actual number of particles formed is only double the initial number of formula units.
Step 3: Using the Formula for Percentage Dissociation
The formula for the percentage dissociation ( ) is given by: Where:
is the van't Hoff factor (2 in this case), is the degree of dissociation, is the number of ions produced per formula unit of solute (which is 4 for MX ).

Step 4: Calculating the Percentage Dissociation
The percentage dissociation is given by:
To the nearest integer, the percentage dissociation is 33%.

To solve this question, we must assess the correctness of both statements in light of the principles of chemistry involved in molal depression constants.

1. Statement (I): Molal depression constant is given by , where:
   • = molar mass of the solvent
   • = universal gas constant
   • = freezing point of the pure solvent
   • = entropy change of fusion
2. Statement (II): for benzene is less than the for water.
   • Typically, the molal depression constant, , depends on the specific properties of the substance, such as its latent heat of fusion.
   • The value for water is known to be approximately 1.86 K kg/mol, whereas for benzene, it is about 5.12 K kg/mol. This means the for benzene is higher, not lower, than that for water.

Now, based on the above evaluation:

• Statement I is indeed correct.
• Statement II is incorrect.

Thus, the most appropriate answer is: Statement I is correct but Statement II is incorrect.

- Statement (I): The addition of NaCl to ice helps lower the freezing point of water, preventing the ice from melting at and thus allowing for the ice cream to stay frozen at lower temperatures. This statement is correct.
- Statement (II): Adding NaCl to ice creates a phenomenon known as freezing point depression, which lowers the freezing point of water and ice. This is a well-known colligative property of solutions. Hence, this statement is also correct.
Therefore, both statements are true.

To understand the graphical representation for the depression in the freezing point of a solvent in the solution, it's essential to understand the concept of colligative properties. Depresssion in the freezing point is one such colligative property, exhibiting the effect of solute addition on the solvent properties. The vapor pressure of the solvent decreases upon solute addition, resulting in a lower freezing point.

1. Concept Explanation:
   • When a non-volatile solute is added to a solvent, the solute particles disrupt the regular crystal structure of the solvent, lowering its freezing point compared to the pure solvent. This phenomenon is known as freezing point depression ( ).
   • In terms of vapor pressure, the presence of a solute decreases the vapor pressure of the solvent. At the freezing point, the solid and liquid phases are in equilibrium.
   • The change in the vapor pressure at different temperatures will show differences from the ideal conditions, causing a shift in the freezing point.
2. Graphical Representation:
   • We need to find the graph showing a drop in the freezing point due to the addition of solute. Look for a graph where the vapor pressure curve intersects at a lower temperature for the solution compared to the pure solvent.
3. Analyzing the Options:
   • The correct graph will show the solution's vapor pressure as being lower than that of the pure solvent, such that the intersection with the solid line occurs at a lower temperature, aligning with the decrease in freezing point ( ).
   • Let's identify such a graph. Based on the provided options, the correct representation is:
4. Conclusion:
   • This graph shows how the vapor pressure curve for the solution intersects the line representing the solid state at a lower temperature when compared to the pure solvent. Thus, it correctly depicts the depression in freezing point ( ) due to the addition of a solute.

To solve this problem, we need to understand how the boiling point of solutions is affected by their concentration and the nature of the solute. The boiling point elevation is a colligative property, which means it depends on the number of solute particles in a solution. The formula for boiling point elevation is:

ΔT_b=iK_bm

Where:

• ΔT_b is the boiling point elevation,
• i is the van't Hoff factor (ion count per molecule),
• K_b is the ebullioscopic constant of the solvent,
• m is the molality of the solution.

Let's analyze each solution:

• (i) 10^-4 M NaCl: NaCl dissociates into 2 ions (Na⁺, Cl^-), so i=2. Molality directly correlates to molarity due to its dilute solution.
• (ii) 10^-4 M urea: Urea does not dissociate; i=1.
• (iii) 10^-3 M NaCl: Higher concentration than (i) with i=2.
• (iv) 10^-2 M NaCl: Highest concentration with i=2.

Order these solutions by their boiling point elevation. Higher the product of i and concentration, higher the boiling point:

Resulting order of increasing boiling points:
(ii) < (i) < (iii) < (iv)

The problem involves the ionization of a coordination complex and the depression in freezing point to deduce the correct complex structure.

1. Given that the depression in freezing point ( ) is 0.558°C for a 0.1 molal solution, we can use the formula for freezing point depression: where is the Van't Hoff factor, is the cryoscopic constant (for water ), and is the molality.
2. The value of can be calculated since: . This means the complex dissociates into 3 ions.
3. According to the coordination number of Cr being 6, we evaluate each of the provided complex structures:
   • [Cr(NH ) ]Cl dissociates into 1 complex ion and 3 Cl ions, resulting in 4 ions total. This does not match .
   • [Cr(NH ) ]Cl Cl dissociates into 1 complex ion and 3 Cl ions, resulting in 4 ions total. This does not match .
   • [Cr(NH ) ]Cl dissociates into 1 complex ion and 2 Cl ions, resulting in 3 ions total. This matches .
   • [Cr(NH ) ]Cl dissociates into 1 complex ion and 3 Cl ions, resulting in 4 ions total. This does not match .
4. Thus, the correct structure of the complex is [Cr(NH ) ]Cl , which is consistent with the value of from the empirical data.
5. Therefore, the correct answer is [Cr(NH ) ]Cl .