abnormal-colligative-property-and-vant-hoff-factor

Cluster 0 (10)

Q1.

We have three aqueous solutions of \(\mathrm{NaCl}\) labelled as ' \(\mathrm{A}\) ', ' \(\mathrm{B}\) ' and ' \(\mathrm{C}\) ' with concentration \(0.1 \mathrm{M}\), \(0.01 \mathrm{M}\) and \(0.001 \mathrm{M}\), respectively. The value of van 't Hoff factor(i) for these solutions will be in the order :

\(\mathrm{i}_{\mathrm{A}}<\mathrm{i}_{\mathrm{C}}<\mathrm{i}_{\mathrm{B}}\)
\(\mathrm{i}_{\mathrm{A}}<\mathrm{i}_{\mathrm{B}}<\mathrm{i}_{\mathrm{C}}\)
\(\mathrm{i}_{\mathrm{A}}>\mathrm{i}_{\mathrm{B}}>\mathrm{i}_{\mathrm{C}}\)
\(\mathrm{i}_{\mathrm{A}}=\mathrm{i}_{\mathrm{B}}=\mathrm{i}_{\mathrm{C}}\)

Salt	Values of i (for different conc. of a Salt)
NaCl	0.1 M	0.01 M	0.001 M
	1.87	1.94	1.94

The van 't Hoff factor (i) is used to describe the number of particles a solute formula unit produces in a solution. For an electrolyte like \(\mathrm{NaCl}\), which dissociates completely in very dilute solutions, the theoretical value of \(i\) is approximately 2, since \(\mathrm{NaCl}\) dissociates into \(\mathrm{Na}^+\) and \(\mathrm{Cl}^-\) ions.

In real scenarios, as the concentration of the solution decreases (making the solution more dilute), the interaction between the ions decreases, allowing more complete dissociation. Therefore, for practical purposes, the van 't Hoff factor \(i\) approaches its theoretical maximum value as concentration decreases. Thus, for \(\mathrm{NaCl}\) solutions of concentrations \(0.1 \mathrm{M}\), \(0.01 \mathrm{M}\), and \(0.001 \mathrm{M}\), the fact that \(\mathrm{NaCl}\) dissociates more completely in more dilute solutions implies that \(i\) increases with decreasing concentration.

Hence, the order of \(i\) based on the concentration would be \(\mathrm{i}_{\mathrm{A}} < \mathrm{i}_{\mathrm{B}} < \mathrm{i}_{\mathrm{C}}\), as concentration \(0.1 \mathrm{M} > 0.01 \mathrm{M} > 0.001 \mathrm{M}\), respectively. Thus, option B correctly describes the order of the van 't Hoff factors for these solutions.

Q2.

\(20 \%\) of acetic acid is dissociated when its \(5 \mathrm{~g}\) is added to \(500 \mathrm{~mL}\) of water. The depression in freezing point of such water is _________ \(\times 10^{-3}{ }^{\circ} \mathrm{C}\).

Atomic mass of \(\mathrm{C}, \mathrm{H}\) and \(\mathrm{O}\) are 12,1 and 16 a.m.u. respectively.

[Given : Molal depression constant and density of water are \(1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\) and \(1 \mathrm{~g} \mathrm{~cm}^{-3}\) respectively.]

\(\begin{aligned} & \mathrm{i}=1+(\mathrm{n}-1) \alpha \\\\ & \Rightarrow \mathrm{i}=1+0.2(2-1)=1.2 \\\\ & \Delta \mathrm{T}_{\mathrm{f}}=\mathrm{i} \mathrm{K} \mathrm{K}_{\mathrm{f}} \mathrm{m} \\\\ & \Delta \mathrm{T}_{\mathrm{f}}=1.2 \times 1.86 \times \frac{5 \times 1000}{60 \times 500} \\\\ & \Delta \mathrm{t}_{\mathrm{f}}=0.372 \\\\ & \Delta \mathrm{T}_{\mathrm{f}}=372 \times 10^{-3}\end{aligned}\)

Q3.

25 mL of an aqueous solution of KCl was found to require 20 mL of 1 M \(\mathrm{AgNO_3}\) solution when titrated using \(\mathrm{K_2CrO_4}\) as an indicator. What is the depression in freezing point of KCl solution of the given concentration? _________ (Nearest integer).

(Given : \(\mathrm{K_f=2.0~K~kg~mol^{-1}}\))

Assume 1) 100% ionization and 2) density of the aqueous solution as 1 g mL\(^{-1}\)

\(25 \times M=20 \times 1\)

\( \begin{aligned} M & =\frac{20}{25}=\frac{4}{5}=0.8 \\\\ \Delta T_{f} & =(\mathrm{i})\left(\mathrm{K}_{\mathrm{f}}\right)(\mathrm{m}) \\\\ & =(2)(2)\left(\frac{4}{5}\right)=\frac{16}{5}=3.2 \end{aligned} \)

Q4.

Lead storage battery contains \(38 \%\) by weight solution of \(\mathrm{H}_{2} \mathrm{SO}_{4}\). The van't Hoff factor is \(2.67\) at this concentration. The temperature in Kelvin at which the solution in the battery will freeze is ________. (Nearest integer).

Given \(\mathrm{K}_{f}=1.8 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\)

\(\mathrm{\Delta T_f=K_f~i~m}\)

\( = 1.8 \times 2.67 \times {{{{38} \over {98}}} \over {0.062}} = 30\)

\(\therefore\) It freeze at \(273-30=243\) \(\mathrm{K}\)

Q5.

1.2 mL of acetic acid is dissolved in water to make 2.0 L of solution. The depression in freezing point observed for this strength of acid is 0.0198\(^\circ\)C. The percentage of dissociation of the acid is ___________. (Nearest integer)

[Given : Density of acetic acid is 1.02 g mL^\(-\)1, Molar mass of acetic acid is 60 g mol^\(-\)1, K_f(H₂O) = 1.85 K kg mol^\(-\)1]

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

Moles of solute \((\) acetic acid \()=\frac{1.2 \times 1.02}{60}\)

As moles of solute are very less.

So, take molarity and molality the same.

\(0.0198=\mathrm{i} \times 1.85 \times \frac{1.2 \times 1.02}{60 \times 2}\)

\(\mathrm{i}=1.05\)

\(\alpha=\frac{i-1}{n-1}=\frac{0.05}{1}= 0.05 = 5\text{%}\)

Q6.

Elevation in boiling point for 1.5 molal solution of glucose in water is 4 K. The depression in freezing point for 4.5 molal solution of glucose in water is 4 K. The ratio of molal elevation constant to molal depression constant (K_b/K_f) is _________.

\(\begin{aligned} &\Delta \mathrm{T}_{\mathrm{b}}=\mathrm{iK}_{\mathrm{b}} \mathrm{m} \\\\ &\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{iK}_{\mathrm{f}} \mathrm{m} \\\\ &\frac{4}{4}=\frac{\mathrm{K}_{\mathrm{b}} 1.5}{\mathrm{~K}_{\mathrm{f}} 4.5} \\\\ &\frac{\mathrm{K}_{\mathrm{b}}}{\mathrm{K}_{\mathrm{f}}}=3 \end{aligned}\)

Q7.

A 0.5 percent solution of potassium chloride was found to freeze at \(-\)0.24\(^\circ\)C. The percentage dissociation of potassium chloride is ______________. (Nearest integer)

(Molal depression constant for water is 1.80 K kg mol^\(-\)1 and molar mass of KCl is 74.6 g mol^\(-\)1)

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

\( \begin{aligned} &\mathrm{i}=\frac{0.24 \times 99.5 \times 74.6}{1.80 \times 0.5 \times 1000} \\\\ &=1.98 \\\\ &\alpha=\frac{\mathrm{i}-1}{\mathrm{n}-1}=\frac{0.98}{1}=0.98 = 98 \,\% \end{aligned} \)

Q8.

Solute A associates in water. When 0.7 g of solute A is dissolved in 42.0 g of water, it depresses the freezing point by 0.2\(^\circ\)C. The percentage association of solute A in water, is :

[Given : Molar mass of A = 93 g mol^\(-\)1. Molal depression constant of water is 1.86 K kg mol^\(-\)1.]

Since, \(\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{ik}_{\mathrm{f} m}\)

\( \begin{aligned} &m=\frac{0.7}{93} \times \frac{1000}{42} \\\\ &0.2=i \times 1.86 \times \frac{0.7 \times 1000}{93 \times 42} \\\\ &i=0.6 \\\\ &\alpha=\frac{i-1}{\frac{1}{n}-1}=\frac{0.6-1}{\frac{1}{2}-1}=0.8 \end{aligned} \)

Hence, the percentage association of solute \(A\) is \(80 \%\).

Q9.

The depression in freezing point observed for a formic acid solution of concentration \(0.5 \mathrm{~mL} \mathrm{~L}^{-1}\) is \(0.0405^{\circ} \mathrm{C}\). Density of formic acid is \(1.05 \mathrm{~g} \mathrm{~mL}^{-1}\). The Van't Hoff factor of the formic acid solution is nearly : (Given for water \(\mathrm{k}_{\mathrm{f}}=1.86\, \mathrm{k} \,\mathrm{kg}\,\mathrm{mol}^{-1}\) )

\(\Delta \mathrm{T}_{\mathrm{f}}\) of formic acid \(=0.0405^{\circ} \mathrm{C}\)

Concentration \(=0.5 \mathrm{~mL} / \mathrm{L}\)

and density \(=1.05 \mathrm{~g} / \mathrm{mL}\)

\(\therefore\) Mass of formic acid in solution \(=1.05 \times 0.5 \mathrm{~g}\)

\( =0.525 \mathrm{~g} \)

\(\therefore\) According to Van't Hoff equation,

\( \begin{aligned} &\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{i} \mathrm{k}_{\mathrm{f}} \cdot \mathrm{m} \\ &0.0405=\mathrm{i} \times 1.86 \times \frac{0.525}{46 \times 1} \end{aligned} \)

(Assuming mass of \(1 \mathrm{~L}\) water \(=\mathrm{kg}\) )

\( \mathrm{i}=\frac{0.0405 \times 46}{1.86 \times 0.525}=1.89 \approx 1.9 \)

Q10.

Two solutions A and B are prepared by dissolving 1 g of non-volatile solutes X and Y, respectively in 1 kg of water. The ratio of depression in freezing points for A and B is found to be 1 : 4. The ratio of molar masses of X and Y is

1 : 4
1 : 0.25
1 : 0.20
1 : 5

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

\( \frac{\Delta T_{\mathrm{f}(\mathrm{A})}}{\Delta \mathrm{T}_{\mathrm{f}(\mathrm{B})}}=\frac{1}{4} \)

\(\frac{\mathrm{i} \times \mathrm{K}_{\mathrm{f}} \times \frac{1}{\mathrm{M}_{\mathrm{A}}} \times 1}{\mathrm{i} \times \mathrm{K}_{\mathrm{f}} \times \frac{1}{\mathrm{M}_{\mathrm{B}}} \times 1}=\frac{1}{4}\)

\(\frac{M_{B}}{M_{A}}=\frac{1}{4}\)

\(M_{A}: M_{B}=4: 1\)

Cluster 1 (4)

Q1.

Solution of \(12 \mathrm{~g}\) of non-electrolyte (A) prepared by dissolving it in \(1000 \mathrm{~mL}\) of water exerts the same osmotic pressure as that of \(0.05 ~\mathrm{M}\) glucose solution at the same temperature. The empirical formula of \(\mathrm{A}\) is \(\mathrm{CH}_{2} \mathrm{O}\). The molecular mass of \(\mathrm{A}\) is __________ g. (Nearest integer)

To solve this problem, we will first calculate the osmotic pressure of the 0.05 M glucose solution and then use that information to determine the molecular mass of compound A.

1. Osmotic pressure equation:

\(\Pi = iMRT\)

where \(\Pi\) is the osmotic pressure, \(i\) is the van't Hoff factor (which is 1 for non-electrolytes), \(M\) is the molarity, \(R\) is the ideal gas constant (0.0821 L atm/mol K), and \(T\) is the temperature in Kelvin.

Since both solutions have the same osmotic pressure at the same temperature, we can set their osmotic pressures equal to each other:

\(\Pi_{A} = \Pi_{glucose}\)

2. Calculate the osmotic pressure of the 0.05 M glucose solution:

Glucose is a non-electrolyte, so its van't Hoff factor is 1. We don't know the temperature, but since both solutions are at the same temperature, it will cancel out in our calculations.

\(\Pi_{glucose} = (1)(0.05 ~\mathrm{M})(R)(T)\)

3. Calculate the molarity of compound A:

Since we know that 12 g of compound A is dissolved in 1000 mL of water, we can find the molarity once we know the molecular mass.

Let \(x\) be the molecular mass of compound A. Then, the molarity of compound A is:

\(M_{A} = \frac{12 ~\mathrm{g}}{x ~\mathrm{g/mol}} \times \frac{1}{1 ~\mathrm{L}} = \frac{12}{x} ~\mathrm{M}\)

4. Set the osmotic pressures equal to each other:

\(\Pi_{A} = \Pi_{glucose}\)

\((1)(\frac{12}{x} ~\mathrm{M})(R)(T) = (1)(0.05 ~\mathrm{M})(R)(T)\)

The van't Hoff factors, ideal gas constant, and temperature cancel out:

\(\frac{12}{x} = 0.05\)

5. Solve for the molecular mass (x) of compound A:

\(x = \frac{12}{0.05}\)

\(x = 240 ~\mathrm{g/mol}\)

The molecular mass of compound A is 240 g/mol.

Q2.

0.004 M K\(_2\)SO\(_4\) solution is isotonic with 0.01 M glucose solution. Percentage dissociation of K\(_2\)SO\(_4\) is ___________ (Nearest integer)

Given that isotonic solutions have the same osmotic pressure, we equate the osmotic pressures of the K₂SO₄ and glucose solutions :

\(i \times 0.004 \, \text{M} = 0.01 \, \text{M}\)

Here, \(i\) is the van't Hoff factor, which accounts for the number of ions the compound dissociates into in the solution. Solving this equation for \(i\), we get \(i = 2.5\).

For the compound K₂SO₄, which dissociates into 3 ions (2K⁺ and 1 SO₄^2-), the formula for \(i\) is given by

\(i = 1 + (n-1)\alpha\)

where \(n\) is the number of ions the solute dissociates into and \(\alpha\) is the degree of dissociation.

Substituting \(n=3\) and \(i=2.5\) into this formula and solving for \(\alpha\) gives

\(\alpha = \frac{i-1}{n-1} = \frac{2.5-1}{3-1} = 0.75\)

To convert this into a percentage, we multiply by 100, yielding a percentage dissociation of 75%.

Q3.

2.5 g of protein containing only glycine (C₂H₅NO₂) is dissolved in water to make 500 mL of solution. The osmotic pressure of this solution at 300 K is found to be 5.03 \(\times\) 10^\(-\)3 bar. The total number of glycine units present in the protein is ____________.

(Given : R = 0.083 L bar K^\(-\)1 mol^\(-\)1)

Since,

\( \pi=\mathrm{icR} \mathrm{T} \)

\(5.03 \times 10^{-3}=\frac{2.5}{M} \times \frac{1000}{500} \times 0.083 \times 300\)

Molar mass of protein \(=24751.5 \mathrm{~g} / \mathrm{mol}\)

Number of glycine units in protein \(=\frac{24751.5}{75}\)

\( =330 \)

Q4.

The osmotic pressure exerted by a solution prepared by dissolving 2.0 g of protein of molar mass 60 kg mol^\(-\)1 in 200 mL of water at 27\(^\circ\)C is ______________ Pa. [integer value]

(use R = 0.083 L bar mol^\(-\)1 K^\(-\)1)

\(\pi=\mathrm{i} C R T \quad(\mathrm{i}=1)\)

\(\pi=\frac{2 \times 1000}{60 \times 10^{3} \times 200} \times .083 \times 300\)

\(\pi=0.00415 \mathrm{~atm}\)

\(\pi=415 \mathrm{~Pa}\)

Noise (7)

Q1.

If A₂B is 30% ionised in an aqueous solution, then the value of van't Hoff factor (i) is _______ × 10⁻¹.

Percent ionisation of \(A_2B=30\%\)

The dissociation of \(A_2B\) in aqueous solution can be represented as

\(A_2B \rightarrow 2A^+ + B^{2-}\)

1 mole of \(A_2B\) produces 2 moles of \(A^+\) and 1 mole of \(B^{2-}\) ions.

The total number of moles of ions produced (n) from the dissociation is

\(n=2+1=3\)

The degree of dissociation given that \(A_2B\) is 30% ionised, the degree of dissociation \((\lambda)\) or degree of ionisation

\(\lambda=\frac{30}{100}=0.3\)

Van't Hoff factor (\(i\)) can be calculated using the formula,

\(i=1+(n-1)\lambda\)

Substitute the values of \(n\) and \(\lambda\) as,

\(i = 1 + (3 - 1) \times 0.3\)

\( = 1 + 2 \times 0.3\)

\( = 1 + 0.6\)

\( = 1.6\)

\( = 16 \times {10^{ - 1}}\)

Q2.

Evaluate the following statements for their correctness.

A. The elevation in boiling point temperature of water will be same for \(0.1 \mathrm{M} \, \mathrm{NaCl}\) and \(0.1 \mathrm{M}\) urea.

B. Azeotropic mixtures boil without change in their composition.

C. Osmosis always takes place from hypertonic to hypotonic solution.

D. The density of \(32 \% \, \mathrm{H}_{2} \mathrm{SO}_{4}\) solution having molarity \(4.09 ~\mathrm{M}\) is approximately \(1.26 \mathrm{~g} \mathrm{~mL}^{-1}\)

E. A negatively charged sol is obtained when KI solution is added to silver nitrate solution.

Choose the correct answer from the options given below :

A, B and D only
A and C only
B and D only
B, D and E only

(A) Elevation in boiling point temperature of water will be higher for 0.1 M NaCl as compared to 0.1 M urea.

(B) Azeotropic mixtures boil without change in their composition

(C) Osmosis always takes place from hypotonic (low concentration of solute) solution to hypertonic (high concentration of solute) solution.

(E) When KI solution is added to AgNO₃ solution, positively charged solution is formed due to adsorption of Ag⁺ ions from dispersion medium.

AgI/ Ag⁺ (positively charged)

(D) Let the mass of H₂SO₄ (32%) is 100.

\( \therefore \text { Weight of } \mathrm{H}_2 \mathrm{SO}_4=32 \)

Moles of \(\mathrm{H}_2 \mathrm{SO}_4=\frac{32}{98}\)

Now, \(4.09=\frac{32}{98 \times \mathrm{V}} \Rightarrow \mathrm{V}=79 \mathrm{ml}\)

Density \(=\frac{100}{79}=1.265\)

Hence, correct answer is (C) B and D only.

Q3.

Match List I with List II

	List I		List II
A.	van't Hoff factor, i	I.	Cryoscopic constant
B.	\(\mathrm{k_f}\)	II.	Isotonic solutions
C.	Solutions with same osmotic pressure	III.	\(\mathrm{\frac{Normal\,molar\,mass}{Abnormal\,molar\,mass}}\)
D.	Azeotropes	IV.	Solutions with same composition of vapour above it

Choose the correct answer from the options given below :

A-III, B-I, C-IV, D-II
A-III, B-I, C-II, D-IV
A-III, B-II, C-I, D-IV
A-I, B-III, C-II, D-IV

A.	van't Hoff factor	III.	\(\mathrm{{{Normal\,molar\,mass} \over {Abnormal\,molar\,mass}}}\)
B.	\(\mathrm{k_f}\)	I.	Cryoscopic constant
C.	Solutions with same osmotic pressure	II.	Isotonic solutions
D.	Azeotropes	IV.	Solutions with same composition of vapour above it

Q4.

Solid Lead nitrate is dissolved in 1 litre of water. The solution was found to boil at 100.15\(^\circ\)C. When 0.2 mol of NaCl is added to the resulting solution, it was observed that the solution froze at \(-0.8^\circ\) C. The solubility product of PbCl\(_2\) formed is __________ \(\times\) 10\(^{-6}\) at 298 K. (Nearest integer)

Given : \(\mathrm{K_b=0.5}\) K kg mol\(^{-1}\) and \(\mathrm{K_f=1.8}\) K kg mol\(^{-1}\). Assume molality to the equal to molarity in all cases.

\(0.15=3 \times 0.5 \times \mathrm{M}\)

\( \begin{aligned} & \mathrm{M}_{\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}}=0.1 \text { molar } \\\\ & \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}+2 \mathrm{NaCl} \longrightarrow \mathrm{PbCl}_{2}+2 \mathrm{NaNO}_{3} \\\\ & \Delta \mathrm{T}_{\mathrm{f}}=\mathrm{iK}_{\mathrm{f}} \cdot \mathrm{m} \\\\ & 0.8=(0.4+3 s) 1.8 \\\\ & s=0.0148 \\\\ & \therefore \text { solubility product }=4 \mathrm{~s}^{3} \\\\ & =4 \times(0.0148)^{3} \\\\ & \approx 13 \times 10^{-6} \end{aligned} \)

Q5.

The number of pairs of the solutions having the same value of the osmotic pressure from the following is _________.

(Assume 100% ionization)

A. 0.500 \(\mathrm{M~C_2H_5OH~(aq)}\) and 0.25 \(\mathrm{M~KBr~(aq)}\)

B. 0.100 \(\mathrm{M~K_4[Fe(CN)_6]~(aq)}\) and 0.100 \(\mathrm{M~FeSO_4(NH_4)_2SO_4~(aq)}\)

C. 0.05 \(\mathrm{M~K_4[Fe(CN)_6]~(aq)}\) and 0.25 \(\mathrm{M~NaCl~(aq)}\)

D. 0.15 \(\mathrm{M~NaCl~(aq)}\) and 0.1 \(\mathrm{M~BaCl_2~(aq)}\)

E. 0.02 \(\mathrm{M~KCl.MgCl_2.6H_2O~(aq)}\) and 0.05 \(\mathrm{M~KCl~(aq)}\)

\(\pi=i C R T\)

The following pairs of solutions have same value of osmotic pressure

(A) \(0.500 \mathrm{M} ~\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\mathrm{aq}) \mathrm{i}=1\) and \(0.25 \mathrm{M} ~\mathrm{KBr}(\mathrm{aq})\) \(i=2\)

\( \begin{aligned} & \pi_1=0.5 \times 1 \times R T =0.5 R T \\\\ \mathrm{KBr} & \rightleftharpoons \mathrm{K}^{+}+\mathrm{Br}^{-} \\\\ & \pi_2=0.25 \times 2 \times R T=0.5 R T \end{aligned} \)

Thus, \(\pi_1=\pi_2\)

(B) \(0.100 \mathrm{M} ~\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right](\mathrm{aq}) \mathrm{i}=5\) and \(0.100 \mathrm{M}\) \(\mathrm{FeSO}_{4}\left(\mathrm{NH}_{4}\right)_{2}(\mathrm{aq}) \mathrm{i}=5\)

\( \begin{array}{ll} \mathrm{K}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right] \rightleftharpoons 4 \mathrm{~K}^{+}+\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-} (i=5) \\\\ \quad \pi_1=0.1 \times 5 \times R T=0.5 R T & \\\\ \mathrm{FeSO}_4 \cdot\left(\mathrm{NH}_4\right)_2 \mathrm{SO}_4 \rightleftharpoons \mathrm{Fe}^{2+}+2 \mathrm{NH}_4^{+}+2 \mathrm{SO}_4^{2-} & (i=5) \\\\ \pi_2=0.1 \times 5 \times R T=0.5 R T & \end{array} \)

Thus, \(\pi_1=\pi_2\)

C. 0.05 \(\mathrm{M~K_4[Fe(CN)_6]~(aq)}\) and 0.25 \(\mathrm{M~NaCl~(aq)}\)

\( \begin{gathered} {K}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right] \rightleftharpoons 4 {~K}^{+}+\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-} (i=5) \\\\ \pi_1=0.05 \times 5 \times R T=0.25 R T \\\\ {NaCl} \rightleftharpoons \mathrm{Na}^{+}+\mathrm{Cl}^{-} \quad(i=2) \\\\ \pi_2=0.25 \times 2 R T=0.5 R T \end{gathered} \)

Thus, \(\pi_1 \neq \pi_2\)

(D) \(0.15 \mathrm{M}~ \mathrm{NaCl}(\mathrm{aq}) \mathrm{i}=2\) and \(0.10 \mathrm{M}~ \mathrm{BaCl}_{2}\) (aq) \(i=3\)

\( \pi_1=0.15 \times 2 \times R T=0.3 R T\)

\( \begin{aligned} \mathrm{BaCl}_2 \rightleftharpoons \mathrm{Ba}^{2+}+2 \mathrm{Cl}^{-}(i=3) \\\\ \pi_2=0.1 \times 3 \times R T=0.3 R T \end{aligned} \)

Thus, \(\pi_1=\pi_2\)

(E) \(0.02 \mathrm{M}~ \mathrm{KCl} . \mathrm{MgCl}_{2} .6 \mathrm{H}_{2} \mathrm{O}(\mathrm{aq}) \mathrm{i}=5\) and \(0.05 \mathrm{M}\) \(\mathrm{KCl}(\mathrm{aq}) \mathrm{i}=2\)

\( \mathrm{KCl} \cdot \mathrm{MgCl}_2 \cdot 6 \mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{K}^{+}+3 \mathrm{Cl}^{-}+\mathrm{Mg}^{2+} \quad(i=5)\)

\( \begin{aligned} & \pi_1=5 \times 0.02 R T=0.1 R T \\\\ & 0.05 \mathrm{M} \mathrm{KCl}, \mathrm{KCl} \rightleftharpoons \mathrm{K}^{+}+\mathrm{Cl}^{-}(i=2) \\\\ & \pi_2=0.05 \times 2 R T=0.1 R T \end{aligned} \)

Thus, \(\pi_1=\pi_2\)

Q6.

If the degree of dissociation of aqueous solution of weak monobasic acid is determined to be 0.3, then the observed freezing point will be ___________% higher than the expected/theoretical freezing point. (Nearest integer)

The degree of dissociation, often represented as \(\alpha\), is the fraction of a mole of a substance that has dissociated into ions in solution. For a weak monobasic acid, this degree of dissociation can increase the number of particles in solution, which can in turn affect colligative properties such as the freezing point.

In this case, a weak monobasic acid, when it dissociates, produces two particles: one \(H^{+}\) ion and one anion. So if the degree of dissociation is 0.3 (\(\alpha = 0.3\)), the average number of particles per molecule of the acid (\(i\)), also known as the van't Hoff factor, will be \(1 + \alpha = 1 + 0.3 = 1.3\).

The decrease in freezing point (\(\Delta T_{f}\)) is given by the formula \(\Delta T_{f} = i \cdot m \cdot K_{f}\), where \(m\) is the molality of the solution and \(K_{f}\) is the cryoscopic constant of the solvent. Therefore, the observed freezing point depression will be 1.3 times the theoretical freezing point depression for a non-dissociating solute (where \(i = 1\)).

So, the observed freezing point will be 30% higher than the theoretical freezing point, given the degree of dissociation of 0.3.

Q7.

2 g of a non-volatile non-electrolyte solute is dissolved in 200 g of two different solvents A and B whose ebullioscopic constants are in the ratio of 1 : 8. The elevation in boiling points of A and B are in the ratio \({x \over y}\) (x : y). The value of y is ______________. (Nearest integer)

\(\Delta T b=k b m\)

\( \begin{aligned} &\frac{\left(\Delta T_{b}\right)_{A}}{\left(\Delta T_{b}\right)_{B}}=\frac{\left(k_{b}\right)_{A}}{\left(k_{b}\right)_{B}} \\\\ &=\frac{1}{8}=\frac{x}{y} \\\\ &\therefore y=8 \end{aligned} \)

Cluster Summary

Cluster 0 (10)

Cluster 1 (4)

Noise (7)