depression-in-freezing-point

Total clusters: 3, Total questions: 16

Cluster Summary

Cluster 0 (3)

Q1.

\(2.7 \mathrm{~kg}\) of each of water and acetic acid are mixed. The freezing point of the solution will be \(-x^{\circ} \mathrm{C}\). Consider the acetic acid does not dimerise in water, nor dissociates in water. \(x=\) ________ (nearest integer)

[Given: Molar mass of water \(=18 \mathrm{~g} \mathrm{~mol}^{-1}\), acetic acid \(=60 \mathrm{~g} \mathrm{~mol}^{-1}\)

\({ }^{\mathrm{K}_{\mathrm{f}}} \mathrm{H}_2 \mathrm{O}: 1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\)

\(\mathrm{K}_{\mathrm{f}}\) acetic acid: \(3.90 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\)

freezing point: \(\mathrm{H}_2 \mathrm{O}=273 \mathrm{~K}\), acetic acid \(=290 \mathrm{~K}\)]

\(\begin{aligned} & \text { Molality of acetic acid }=\frac{2700}{60} \times \frac{1}{2.7} \mathrm{~mol} / \mathrm{kg} \\ &=16.667 \\ & \Delta \mathrm{T}_{\mathrm{f}}=\mathrm{K}_{\mathrm{f}} \times 16.667 \\ &=1.86 \times 16.667 \\ &=31 \mathrm{~K} \end{aligned}\)

Q2.

Considering acetic acid dissociates in water, its dissociation constant is \(6.25 \times 10^{-5}\). If \(5 \mathrm{~mL}\) of acetic acid is dissolved in 1 litre water, the solution will freeze at \(-x \times 10^{-2}{ }^{\circ} \mathrm{C}\), provided pure water freezes at \(0{ }^{\circ} \mathrm{C}\).

\(x=\) _________. (Nearest integer)

\(\begin{aligned} \text{Given :} \quad & \left(\mathrm{K}_{\mathrm{f}}\right)_{\text {water }}=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}-1 \\ & \text { density of acetic acid is } 1.2 \mathrm{~g} \mathrm{~mol}^{-1} \text {. } \\ & \text { molar mass of water }=18 \mathrm{~g} \mathrm{~mol}^{-1} \text {. } \\ & \text { molar mass of acetic acid= } 60 \mathrm{~g} \mathrm{~mol}^{-1} \text {. } \\ & \text { density of water }=1 \mathrm{~g} \mathrm{~cm}^{-3} \end{aligned}\)

Acetic acid dissociates as \(\mathrm{CH}_3 \mathrm{COOH} \rightleftharpoons \mathrm{CH}_3 \mathrm{COO}^{\ominus}+\mathrm{H}^{\oplus}\)

To solve the problem, we'll calculate the freezing point depression of the acetic acid solution in water.

  1. Calculating Moles of Acetic Acid:
  • Given volume of acetic acid: \( 5 \, \text{mL} \)

  • Density of acetic acid: \( 1.2 \, \text{g/mL} \)

  • Molar mass of acetic acid: \( 60 \, \text{g/mol} \)

\( \text{Mass of acetic acid} = \text{Volume} \times \text{Density} = 5 \, \text{mL} \times 1.2 \, \text{g/mL} = 6 \, \text{g} \)

\( \text{Moles of acetic acid} = \frac{\text{Mass}}{\text{Molar mass}} = \frac{6 \, \text{g}}{60 \, \text{g/mol}} = 0.1 \, \text{mol} \)

  1. Calculating Molality:
  • Volume of water: \( 1 \, \text{L} \)

  • Density of water: \( 1 \, \text{g/cm}^3 \Rightarrow 1 \, \text{kg/L} \)

  • Hence, mass of water = \( 1 \, \text{kg} \)

\( \text{Molality (m)} = \frac{\text{Moles of solute}}{\text{Mass of solvent (kg)}} = \frac{0.1 \, \text{mol}}{1 \, \text{kg}} = 0.1 \, \text{mol/kg} \)

  1. Degree of Dissociation (\(\alpha\)):
  • Dissociation constant (\(K_a\)) of acetic acid: \(6.25 \times 10^{-5}\)

\( \alpha = \sqrt{\frac{K_a}{C}} = \sqrt{\frac{6.25 \times 10^{-5}}{0.1}} = \sqrt{6.25 \times 10^{-4}} = 25 \times 10^{-3} = 0.025 \)

  1. Van't Hoff factor (i):
  • Acetic acid partially dissociates into 2 ions (CH\(_3\)COO\(^-\) and H\(^+\))

\( i = 1 + (\text{number of ions produced} - 1) \times \alpha = 1 + (2-1) \times 0.025 = 1 + 0.025 = 1.025 \)

  1. Freezing Point Depression (\(\Delta T_f\)):
  • Freezing point depression constant (\(K_f\)) for water: \(1.86 \,\text{K kg/mol}\)

\( \Delta T_f = i \times K_f \times \text{Molality} = 1.025 \times 1.86 \times 0.1 = 0.19065 \, \text{K} \)

  1. Final Freezing Point:
  • Pure water freezes at \(0 \, ^\circ \text{C}\)

\( \text{Freezing point of solution} = 0 \, ^\circ \text{C} - 0.19065 \, ^\circ \text{C} = -0.19065 \, ^\circ \text{C} \)

Here, \(-x \times 10^{-2} \, ^\circ \text{C}\) is given. Therefore, \(x = 19\) (nearest integer).

So, the final answer is:

\( x = 19 \)

Q3.

\(150 \mathrm{~g}\) of acetic acid was contaminated with \(10.2 \mathrm{~g}\) ascorbic acid \(\left(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{6}\right)\) to lower down its freezing point by \(\left(x \times 10^{-1}\right)^{\circ} \mathrm{C}\). The value of \(x\) is ___________. (Nearest integer)

[Given \(\mathrm{K}_{f}=3.9 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\); molar mass of ascorbic acid \(=176 \mathrm{~g} \mathrm{~mol}^{-1}\)]

M.wt. of Acetic acid \(=60 \mathrm{~g}\)

M.wt. of Ascorbic acid \(=176 \mathrm{~g}\)

\( \Delta T_{f}=K_{f} m \)

\( \begin{aligned} \Delta T_{f} &=\frac{3.9 \times 10.2 \times 1000}{176 \times 150} \\\\ \Delta T_{f} &=1.506 \\\\ &=15.06 \times 10^{-1} \\\\ &=15 \end{aligned} \)

Cluster 1 (4)

Q1.

What is the freezing point depression constant of a solvent, 50 g of which contain 1 g non volatile solute (molar mass \(256 \mathrm{~g} \mathrm{~mol}^{-1}\) ) and the decrease in freezing point is 0.40 K ?

  1. \(4.43 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\)
  2. \(3.72 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\)
  3. \(5.12 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\)
  4. \(1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\)

To find the freezing point depression constant (\( K_f \)) of the solvent, we use the formula for freezing point depression:

\( \Delta T_f = K_f \cdot m \)

Given:

The decrease in freezing point \(\Delta T_f\) is 0.40 K.

The mass of the solute is 1 g and its molar mass is 256 g/mol.

The mass of the solvent is 50 g (or 0.050 kg).

First, calculate the molality (\( m \)):

Molality is defined as the moles of solute per kilogram of solvent.

Calculate moles of solute:

\( \text{Moles of solute} = \frac{1 \, \text{g}}{256 \, \text{g/mol}} = \frac{1}{256} \, \text{mol} \)

Calculate molality (\( m \)):

\( m = \frac{\frac{1}{256} \, \text{mol}}{0.050 \, \text{kg}} = \frac{1}{256 \times 0.050} \, \text{mol/kg} \)

Now, substitute into the formula to find \( K_f \):

\( 0.4 = K_f \cdot \frac{1}{256 \times 0.050} \)

Solving for \( K_f \):

\( K_f = 0.4 \cdot 256 \times 0.050 = 5.12 \, \text{K kg/mol} \)

Thus, the freezing point depression constant of the solvent is \( 5.12 \, \text{K kg/mol} \).

Q2.
\(\mathrm{HA}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{A}^{-}(a q)\)

The freezing point depression of a 0.1 m aqueous solution of a monobasic weak acid HA is 0.20 °C. The dissociation constant for the acid is

Given: \(K_f\)(H2O) = 1.8 K kg mol−1, molality ≡ molarity

  1. \(1.90 \times 10^{-3}\)

  2. \(1.38 \times 10^{-3}\)

  3. \(1.1 \times 10^{-2}\)

  4. \(1.89 \times 10^{-1}\)

Freezing Point Depression:

\( \Delta T_f = i \times K_f \times m \)

Given:

\( \Delta T_f = 0.2 \, \text{°C}, \quad K_f = 1.8 \, \text{K kg mol}^{-1}, \quad m = 0.1 \, \text{m} \)

Substituting the given values:

\( 0.2 = i \times 1.8 \times 0.1 \)

Solving for \(i\):

\( i = \frac{0.2}{1.8 \times 0.1} = \frac{20}{18} = \frac{10}{9} \)

Degree of Dissociation (\(\alpha\)):

For the reaction \(\mathrm{HA} \rightleftharpoons \mathrm{H}^{+} + \mathrm{A}^{-}\):

\( i = 1 + \alpha \)

Given \(i = \frac{10}{9}\):

\( \frac{10}{9} = 1 + \alpha \)

\( \alpha = \frac{1}{9} \)

Dissociation Constant (\(K_{eq}\)):

\( \mathrm{K}_{eq} = \frac{[H^+][A^-]}{[HA]} \)

At equilibrium:

\( [H^+] = [A^-] = \alpha \times C = \frac{1}{9} \times 0.1 \)

\( [HA] = 0.1 \times (1 - \alpha) = 0.1 \times \left(1 - \frac{1}{9}\right) \)

Substituting these into \(K_{eq}\):

\( \mathrm{K}_{eq} = \frac{(0.1 \times \frac{1}{9})^2}{0.1 \times \left(1 - \frac{1}{9}\right)} \)

Simplifying:

\( \mathrm{K}_{eq} = \frac{0.1 \times \left(\frac{1}{81}\right)}{0.1 \times \frac{8}{9}} = \frac{1}{720} \)

Therefore:

\( \mathrm{K}_{eq} = 1.38 \times 10^{-3} \)

Q3.

The elevation in boiling point for 1 molal solution of non-volatile solute A is \(3 \mathrm{~K}\). The depression in freezing point for 2 molal solution of \(\mathrm{A}\) in the same solvent is 6 \(K\). The ratio of \(K_{b}\) and \(K_{f}\) i.e., \(K_{b} / K_{f}\) is \(1: X\). The value of \(X\) is [nearest integer]

Molality of a solution of non volatile solute \((A)=1\)

Elevation in boiling point is given by

\(\Delta \mathrm{T}_{\mathrm{b}}=\mathrm{K}_{\mathrm{b}} \mathrm{m}\)

\( 3=\mathrm{K}_{\mathrm{b}} \times 1 \)       ... (1)

Molality of \((A)\) in the same solvent \(=2\)

Depression in freezing point is given by

\( \begin{aligned} &\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{K}_{\mathrm{f}} \mathrm{m} \\ &6=\mathrm{K}_{\mathrm{f}} \times 2 \quad\quad\quad{...(2)} \\ &\text { Dividing (1) by (2) } \\ &\frac{\mathrm{K}_{\mathrm{b}}}{\mathrm{K}_{\mathrm{f}}}=\frac{1}{\mathrm{X}}=\frac{1}{1} \\ &\therefore \quad \mathrm{X}=1 \end{aligned} \)
Q4.

\(1.80 \mathrm{~g}\) of solute A was dissolved in \(62.5 \mathrm{~cm}^{3}\) of ethanol and freezing point of the solution was found to be \(155.1 \mathrm{~K}\). The molar mass of solute A is ________ g \(\mathrm{mol}^{-1}\).

[Given : Freezing point of ethanol is 156.0 K.

Density of ethanol is 0.80 g cm\(-\)3.

Freezing point depression constant of ethanol is 2.00 K kg mol\(-\)1]

Mass of solvent \(=d x v=0.8 \times 62.5=50\) gram

\(\Delta T_{f}=k_{f} \times m\)

\( \begin{aligned} &0.9=2\left[\frac{1.8 \times 1000}{M_{\text {Solute }} \times 50}\right] \\ &M_{\text {Solute }}=\left(\frac{2 \times 1.8 \times 1000}{0.9 \times 50}\right)=80 \end{aligned} \)

Noise (9)

Q1.

Given below are two statements :

Statement (I): NaCl is added to the ice at 0°C, present in the ice cream box to prevent the melting of ice cream.

Statement (II): On addition of NaCl to ice at 0°C, there is a depression in freezing point.

In the light of the above statements, choose the correct answer from the options given below :

  1. Both Statement I and Statement II are false

  2. Statement I is true but Statement II is false

  3. Both Statement I and Statement II are true

  4. Statement I is false but Statement II is true

Statement I : Correct

NaCl addition to ice causes preventing the melting of ice. On adding NaCl to ice, freezing point lowers. This creates a colder mixture, preventing the ice cream from melting.

Melting point of ice is 0\(^\circ\)C. When only ice is used to make ice cream, at 0\(^\circ\)C ice starts melting by absorbing the energy from its environment in the form of heat. Addition of salt to ice while making the cream lowers the freezing point of the ice, allowing it to reach a colder temperature and thus the ice cream mixture freezes properly, So, the salt causes ice to melt at a lower temperature than pure ice.

Statement II : Correct

Decrease in freezing point while addition of NaCl to ice at 0\(^\circ\)C is due to the colligative property depression in freezing point.

So, both the statements are correct.

Both statement I and statement II are true.

Q2.

Consider the given plots of vapour pressure (VP) vs temperature(T/K). Which amongst the following options is correct graphical representation showing \(\Delta \mathrm{T}_{\mathrm{f}}\), depression in the freezing point of a solvent in a solution?

  1. JEE Main 2025 (Online) 24th January Morning Shift Chemistry - Solutions Question 15 English Option 1
  2. JEE Main 2025 (Online) 24th January Morning Shift Chemistry - Solutions Question 15 English Option 2
  3. JEE Main 2025 (Online) 24th January Morning Shift Chemistry - Solutions Question 15 English Option 3
  4. JEE Main 2025 (Online) 24th January Morning Shift Chemistry - Solutions Question 15 English Option 4

On adding non-volatile solute in a solvent, the freezing point of solution decreases.

\(\mathrm{T}_{\mathrm{f}}<\mathrm{T}_{\mathrm{f}}^0\)

F.P. of solution \(<\) F.P. of pure solvent

Also V.P. of solution decreases on adding nonvolatile solute in a solvent.

JEE Main 2025 (Online) 24th January Morning Shift Chemistry - Solutions Question 15 English Explanation

Q3.

Given below are two statements:

Statement (I) : Molal depression constant \(\mathrm{K}_f\) is given by \(\frac{\mathrm{M}_1 \mathrm{RT}_f}{\Delta \mathrm{~S}_{\mathrm{fus}}}\), where symbols have their usual meaning.

Statement (II) : \(\mathrm{K}_f\) for benzene is less than the \(\mathrm{K}_f\) for water.

In the light of the above statements, choose the most appropriate answer from the options given below :

  1. Statement I is correct but Statement II is incorrect
  2. Both Statement I and Statement II are incorrect
  3. Statement I is incorrect but Statement II is correct
  4. Both Statement I and Statement II are correct

Statement-I

Molar depression constant \(\mathrm{k}_f=\frac{\mathrm{M}_1 \mathrm{RT}_f^2}{\Delta \mathrm{H}_{\text {fus }}}\)

\(\begin{aligned} & \mathrm{k}_f=\frac{\mathrm{M}_1 \mathrm{RT}_{\mathrm{f}}}{\left[\frac{\Delta \mathrm{H}_{\mathrm{fus}}}{\mathrm{~T}_{\mathrm{f}}}\right]} \\ & \mathrm{k}_f=\frac{\mathrm{M}_1 \mathrm{RT}_f}{\Delta \mathrm{~S}_{\text {fus }}} \end{aligned}\)

Hence statement-I is correct

but \(\mathrm{k}_{\mathrm{f}}\) for benzene \(=5.12 \frac{{ }^{\circ} \mathrm{C}}{\text { molal }}\)

\(\mathrm{k}_{\mathrm{f}}\) for water \(=1.86 \frac{{ }^{\circ} \mathrm{C}}{\text { molal }}\) Hence statement- II is incorrect

Q4.
Mass of ethylene glycol (antifreeze) to be added to \(18.6 \mathrm{~kg}\) of water to protect the freezing point at \(-24^{\circ} \mathrm{C}\) is ________ \(\mathrm{kg}\) (Molar mass in \(\mathrm{g} ~\mathrm{mol}^{-1}\) for ethylene glycol \(62, \mathrm{~K}_f\) of water \(=1.86 \mathrm{~K}\) \(\mathrm{kg} ~\mathrm{mol}^{-1}\) )

To determine the mass of ethylene glycol (antifreeze) needed to lower the freezing point of water to \(-24^{\circ} \mathrm{C}\), we can use the freezing point depression equation, which is a colligative property given by:

\( \Delta T_f = i \cdot K_f \cdot m \)

where:

  • \(\Delta T_f\) is the depression in the freezing point.
  • \(i\) is the van't Hoff factor, which is the number of particles the solute splits into in solution. For ethylene glycol, which does not dissociate in solution, \(i = 1\).
  • \(K_f\) is the cryoscopic constant (freezing point depression constant) of water, which is given as \(1.86 \mathrm{~K} ~\mathrm{kg} ~\mathrm{mol}^{-1}\).
  • \(m\) is the molality of the solution.

However, molality is defined as the number of moles of solute per kilogram of solvent, so we first need to find the molality that corresponds to the desired freezing point depression:

\( \Delta T_f = -24^{\circ} \mathrm{C} - (0^{\circ} \mathrm{C}) = -24^{\circ} \mathrm{C} \)

Now, we can solve for \(m\):

\( m = \frac{\Delta T_f}{i \cdot K_f} \)

Substituting the given values:

\( m = \frac{-24^{\circ} \mathrm{C}}{1 \cdot 1.86 \mathrm{~K} ~\mathrm{kg} ~\mathrm{mol}^{-1}} \)

Note that \(\mathrm{C}\) and \(\mathrm{K}\) are interchangeable when calculating changes in temperature. Now, we can compute the molality:

\( m = \frac{-24}{1.86} \)

Next, we calculate the molality of the ethylene glycol:

\( m \approx -12.903 \mathrm{~mol} ~\mathrm{kg}^{-1} \)

Keep in mind that molality is always positive, but the negative sign indicates the direction of the temperature change. Since we're calculating the amount needed, we can consider it as 12.903 moles per kilogram of water.

Now we calculate the moles of ethylene glycol required for \(18.6 \mathrm{~kg}\) of water:

\( \text{moles of ethylene glycol} = m \cdot \text{mass of water (in kg)} \) \( \text{moles of ethylene glycol} = 12.903 \mathrm{~mol} ~\mathrm{kg}^{-1} \cdot 18.6 \mathrm{~kg} \) \( \text{moles of ethylene glycol} \approx 240.0158 \mathrm{~mol} \)

Next, we find the mass of ethylene glycol needed using its molar mass:

\( \text{mass of ethylene glycol} = \text{moles of ethylene glycol} \cdot \text{molar mass of ethylene glycol} \)

Given the molar mass of ethylene glycol is \(62 \mathrm{~g} ~\mathrm{mol}^{-1}\), we then convert it to \(\mathrm{kg} ~\mathrm{mol}^{-1}\) by dividing by 1000:

\( \text{molar mass of ethylene glycol} = 0.062 \mathrm{~kg} ~\mathrm{mol}^{-1} \)

Now we can calculate the mass:

\( \text{mass of ethylene glycol} = 240.0158 \mathrm{~mol} \cdot 0.062 \mathrm{~kg} ~\mathrm{mol}^{-1} \) \( \text{mass of ethylene glycol} \approx 14.88098 \mathrm{~kg} \)

Therefore, approximately \(14.881 \mathrm{~kg}\) of ethylene glycol needs to be added to \(18.6 \mathrm{~kg}\) of water to lower the freezing point to \(-24^{\circ} \mathrm{C}\).

Q5.

The solution from the following with highest depression in freezing point/lowest freezing point is

  1. \(180 \mathrm{~g}\) of acetic acid dissolved in benzene
  2. \(180 \mathrm{~g}\) of acetic acid dissolved in water
  3. \(180 \mathrm{~g}\) of benzoic acid dissolved in benzene
  4. \(180 \mathrm{~g}\) of glucose dissolved in water

\(\Delta \mathrm{T}_{\mathrm{f}}\) is maximum when \(\mathrm{i} \times \mathrm{m}\) is maximum.

1) \(\mathrm{m}_1=\frac{180}{60}=3, \mathrm{i}=1+\alpha\)

Hence

\(\Delta \mathrm{T}_{\mathrm{f}}=(1+\alpha) \cdot \mathrm{k}_{\mathrm{f}}=3 \times 1.86=5.58^{\circ} \mathrm{C}(\alpha<<1)\)

2) \(\mathrm{m}_2=\frac{180}{60}=3, \mathrm{i}=0.5, \Delta \mathrm{T}_{\mathrm{f}}=\frac{3}{2} \times \mathrm{k}_{\mathrm{f}}{ }^{\prime}=7.68^{\circ} \mathrm{C}\)

3) \(\mathrm{m}_3=\frac{180}{122}=1.48, \mathrm{i}=0.5, \Delta \mathrm{T}_{\mathrm{f}}=\frac{1.48}{2} \times \mathrm{k}_{\mathrm{f}}{ }^{\prime}=3.8^{\circ} \mathrm{C}\)

4) \(\mathrm{m}_4=\frac{180}{180}=1, \mathrm{i}=1, \Delta \mathrm{T}_{\mathrm{f}}=1 \cdot \mathrm{k}_{\mathrm{f}}{ }^{\prime}=1.86^{\circ} \mathrm{C}\)

As per NCERT, \(\mathrm{k}_{\mathrm{f}}{ }^{\prime}\left(\mathrm{H}_2 \mathrm{O}\right)=1.86 \mathrm{~k} \cdot \mathrm{~kg} \mathrm{~mol}^{-1}\)

\(\mathrm{k}_{\mathrm{f}}{ }^{\prime}(\text { Benzene })=5.12 \mathrm{~k} \cdot \mathrm{~kg} \mathrm{~mol}^{-1}\)

Q6.

What happens to freezing point of benzene when small quantity of napthalene is added to benzene?

  1. Increases
  2. Decreases
  3. Remains unchanged
  4. First decreases and then increases

When a small quantity of naphthalene is added to benzene, the freezing point of benzene decreases. This phenomenon is due to the colligative property known as freezing point depression. The addition of a solute, such as naphthalene, disrupts the orderly arrangement of solvent molecules in the solid phase, thereby lowering the freezing point.

Mathematically, the decrease in freezing point (\(\Delta T_f\)) can be represented by the equation:

\( \Delta T_f = i \cdot K_f \cdot m \)

where:

\(i\) is the van't Hoff factor (which is 1 for naphthalene as it does not dissociate in solution),

\(K_f\) is the cryoscopic constant of the solvent (benzene),

\(m\) is the molality of the solution.

Thus, the correct option is:

Option B: Decreases

Q7.

When '\(x\)' \(\times 10^{-2} \mathrm{~mL}\) methanol (molar mass \(=32 \mathrm{~g}\)' density \(=0.792 \mathrm{~g} / \mathrm{cm}^3\)) is added to \(100 \mathrm{~mL}\). water (density \(=1 \mathrm{~g} / \mathrm{cm}^3\)), the following diagram is obtained.

JEE Main 2024 (Online) 6th April Evening Shift Chemistry - Solutions Question 23 English

\(x=\) ________ (nearest integer).

[Given : Molal freezing point depression constant of water at \(273.15 \mathrm{~K}\) is \(1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\)]

\(\begin{aligned} & \Delta T_f=2.5^{\circ} \mathrm{C} \\ & \Delta T_f=i \times k_f \times m \\ & 2.5=1 \times 1.86 \times \frac{n_B}{0.1} \\ & n_B=\frac{2.5 \times 0.1}{1.86}=0.1344 \mathrm{~mol} \\ & \text { mass of methanol }=0.1344 \times 32=4.3 \mathrm{~g} \\ & d=\frac{m}{v} \\ & v=\frac{m}{d} \\ & v=\frac{4.3}{0.792} \mathrm{~mL} \\ & =5.43 \mathrm{~mL} \\ & =543 \times 10^{-2} \mathrm{~mL} \\ & x=543 \end{aligned}\)

Q8.

In the depression of freezing point experiment

A. Vapour pressure of the solution is less than that of pure solvent

B. Vapour pressure of the solution is more than that of pure solvent

C. Only solute molecules solidify at the freezing point

D. Only solvent molecules solidify at the freezing point

Choose the most appropriate answer from the options given below :

  1. A and C only
  2. A only
  3. A and D only
  4. B and C only
In the depression of freezing point experiment only solvent molecules solidify and vapour pressure of solution decreases as some of the surface area is occupied by solute molecules, so less number of molecules will go in vapour form.
Q9.

Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).

Assertion (A) : At 10\(^\circ\)C, the density of a 5 M solution of KCl [atomic masses of K & Cl are 39 & 35.5 g mol\(-\)1 respectively], is 'x' g ml\(-\)1. The solution is cooled to \(-\)21\(^\circ\)C. The molality of the solution will remain unchanged.

Reason (R) : The molality of a solution does not change with temperature as mass remains unaffected with temperature.

In the light of the above statements, choose the correct answer from the options given below :

  1. Both (A) and (R) are true and (R) is the correct explanation of (A).
  2. Both (A) and (R) are true but (R) is not the correct explanation of (A).
  3. (A) is true but (R) is false.
  4. (A) is false but (R) is true.
Molality and Mass are temperature Independent so on changing temp., molality and mass remain unchanged.