elevation-in-boiling-point

Mass of \(A x_2=1.24 \mathrm{~g}\) (solute)

Molarmass of \(A X_2=124 \mathrm{~g} \mathrm{~mol}^{-1}\)

Mass of water \(=1 \mathrm{~kg}\) (solvent.)

Boiling point of water \(=100^{\circ} \mathrm{C}\)

Boiling point of water after adding solute \(A X_2=100.0156^{\circ} \mathrm{C}\)

Mass of \(A Y_2=25.4 \mathrm{~g}\) (solute)

Molarmass of \(A Y_2=250 \mathrm{~g~mol}^{-1}\)

Mass of water \(=2 \mathrm{~kg}\) (Solvent)

Boiling point of water \(\mp 100^{\circ} \mathrm{C}\)

Boiling point of water after adding solute \(A y_2=100.0260^{\circ} \mathrm{C}\)

\(\mathrm{K}_{\mathrm{b}}\left(\mathrm{H}_2 \mathrm{O}\right)=0.52 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}{ }^{-1}=0.520^{\circ} \mathrm{C} \mathrm{kg} \mathrm{mol}^{-1}\)

The ionisation of \(A x_2\) and \(A Y_2\) can be determined by calculating Van't Hoff factor.

The phenomenon given in the question is elevation in boiling point. The boiling point of solvent increases when another compound (solute) is added to it.

Relation: \(\Delta T_b=K_b \cdot i \cdot m\)

\(\Delta T_b=T_b-T_b^0\)

\(\Delta T_b \rightarrow\) Boiling point elevation

\(T_b \rightarrow\) Boiling point of solution (solvent + solute)

\(T_b^0 \rightarrow\) Boiling point of solvent

\(K_b \rightarrow\) Molal elevation constant

\(i \rightarrow\) Van't Hoff factor

\(m \rightarrow \text { Molality }=\frac{\text { Number of moles }}{\mathrm{kg} \text { of solvent }}, \text { Moles }=\frac{\text { Mass }}{\text { Molarmass }}\)

For \(A X_2\) and \(A Y_2\) (solute), Van't Hoff factor represents how many particles a solute dissociates into when dissolved in a solvent (water).

For non-electrolytes, \(i=1\)

\(\begin{aligned} &\begin{aligned} & A X_2: \\ & \text { Moles }=\frac{\text { mass }}{\text { molarmass }}=\frac{1.249}{124 \mathrm{~g~mol^{-1}}}=0.01 \mathrm{~mol} \\ & \text { Molality }=\frac{\text { Moles }}{k_g \text { of solvent }}=\frac{0.01 \mathrm{~mol}}{1 \mathrm{~kg}}=0.01 \mathrm{~mol} \mathrm{~kg}^{-1} \\ & \Delta T_b=K_b \times i \times m \\ & i=\frac{\Delta T_b}{k_b \times m} \\ &=\frac{T_b-T_b^0}{K_b \times m} \end{aligned}\\ &\text { Substitute values as., } \end{aligned}\)

\(\begin{aligned} i & =\frac{\left(100.0156^{\circ} \mathrm{C}-100^{\circ} \mathrm{C}\right)}{0.52^{\circ} \mathrm{C} \mathrm{~kg} \mathrm{~mol}^{-1} \times 0.01 \mathrm{~mol} \mathrm{~kg}^{-1}} \\ & =\frac{0.0156^\circ \mathrm{C}}{0.52 \times 0.01{ }^{\circ} \mathrm{C}}=3 \end{aligned}\)

\(i=3\) means, there are 3 particles in solution after \(A X_2\) dissolved in water.

\(A x_2 \rightarrow 1 A^{+}+2 X^{-} \quad(1+2=3)\)

\(\begin{aligned} &A x_2 \text { is completely ionised. }\\ &A Y_2: \end{aligned}\)

\(\begin{aligned} &\begin{aligned} \text { Moles } & =\frac{\text { Mass }}{\text { Molarmass }}=\frac{25.4 \mathrm{~g}}{250 \mathrm{g~mol}}=0.1016 \mathrm{~mol} \\ \text { Molality } & =\frac{\text { Moles }}{\mathrm{kgof}^{-1} \text { solvent }}=\frac{0.1016 \mathrm{~mol}}{2 \mathrm{~kg}}=0.050 .8 \mathrm{~mol} \mathrm{~kg}^{-1} \\ \Delta T_b & =k_b \times \mathrm{l}^2 \times \mathrm{m} \\ i & =\frac{\Delta T_b}{k_b \times m} \\ & =\frac{T_b-T_b^0}{k_b \times m} \end{aligned}\\ &\text { Substitute values as, } \end{aligned}\)

\(i = {{(100.0260^\circ C - 100^\circ C)} \over {0.52^\circ C\,kg\,mo{l^{ - 1}} \times 0.0508\,mol\,k{g^{ - 1}}}}\)

\(=\frac{0.0260 ^\circ \mathrm{C}}{0.52 \times 0.0508 ^\circ \mathrm{C}}\)

\(= 0.98\)

\(\approx 1\)

\(i=1\) means \(A Y_2\) is completely unionised.

\(A y_2\) not give ionised particles when dissolved in water.

So, \(A X_2\) is completely ionised and \(A Y_2\) is completely unionised.

Answer: Option 4) \(A x_2\) is fully ionised, \(A Y_2\) is completely unionised.

Step 1: Identify the van’t Hoff factor (\(i\)) for each solute

NaCl dissociates (ideally) into two ions:

\( \mathrm{NaCl} \;\rightarrow\; \mathrm{Na^+} + \mathrm{Cl^-}, \)

so \(i \approx 2.\)

Urea (\(\mathrm{CH_4N_2O}\)) is a non‐electrolyte (does not dissociate), so \(i = 1.\)

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Step 2: Effective molar concentration of particles

The total particle concentration for each solution is approximately \((i \times \text{molarity})\).

(i) \(10^{-4}\,M\) NaCl

\( \text{Effective concentration} \;=\; 2 \times 10^{-4} = 2 \times 10^{-4}. \)

(ii) \(10^{-4}\,M\) Urea

\( \text{Effective concentration} \;=\; 1 \times 10^{-4} = 1 \times 10^{-4}. \)

(iii) \(10^{-3}\,M\) NaCl

\( \text{Effective concentration} \;=\; 2 \times 10^{-3} = 2 \times 10^{-3}. \)

(iv) \(10^{-2}\,M\) NaCl

\( \text{Effective concentration} \;=\; 2 \times 10^{-2} = 2 \times 10^{-2}. \)

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Step 3: Compare to rank the boiling points

A larger total particle concentration (and hence larger colligative effect) corresponds to a higher boiling point. Arrange from lowest to highest:

Lowest: \(10^{-4}\,M\) Urea \(\bigl[1 \times 10^{-4}\bigr]\)

Next: \(10^{-4}\,M\) NaCl \(\bigl[2 \times 10^{-4}\bigr]\)

Next: \(10^{-3}\,M\) NaCl \(\bigl[2 \times 10^{-3}\bigr]\)

Highest: \(10^{-2}\,M\) NaCl \(\bigl[2 \times 10^{-2}\bigr]\)

Hence, in the format \((\text{ii}) < (\text{i}) < (\text{iii}) < (\text{iv})\).

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Final Answer

\( \boxed{\text{(ii) } < \text{(i) } < \text{(iii) } < \text{(iv)}} \quad \text{(Option B)} \)

For AB

\(\begin{aligned} & \Delta \mathrm{T}_{\mathrm{b}}=2.7 \mathrm{~K} \\ & 2.7=1 \times 0.5 \times \mathrm{m} \\ & \mathrm{~m}=\frac{27}{5} \end{aligned}\)

Let molar mass of \(A B=x\).

So \(\frac{1 / x}{15} \times 1000=\frac{27}{5}\)

\(x=12.34\)

For \(\mathrm{AB}_2\)

\(\begin{aligned} & \Delta \mathrm{T}_{\mathrm{b}}=1.5 \mathrm{~K} \\ & 1.5=1 \times 0.5 \times \mathrm{m} \\ & \mathrm{~m}=3 \end{aligned}\)

Let molar mass of \(\mathrm{AB}_2=\mathrm{y}\)

So \(\frac{1 / \mathrm{y}}{15} \times 1000=3\)

\(\begin{aligned} & y=\frac{1000}{45} \\ & y=22.22 \end{aligned}\)

Now let a and b be atomic masses of A and B respectively, then

\(\begin{aligned} & \mathrm{A}+\mathrm{b}=12.34 \quad\text{...... (i)}\\ & \mathrm{~A}+2 \mathrm{~b}=22.22 \quad\text{...... (ii)}\\ & \mathrm{~B}=22.22-12.34=9.88 \end{aligned}\)

Now \(\mathrm{a}=12.34-9.88=2.46\)

\(=24.6 \times 10^{-1}=25 \times 10^{-1}\)

Boiling Point Elevation Formula:

\( \Delta T_b = i_1 \cdot m_1 \cdot K_b + i_2 \cdot m_2 \cdot K_b \)

where:

\(i_1\) and \(i_2\) are the van't Hoff factors for ethylene glycol and glucose, respectively (both are 1 since they do not dissociate in solution).

\(m_1\) and \(m_2\) are the molalities of ethylene glycol and glucose, respectively.

\(K_b\) is the ebullioscopic constant of water (\(0.52 \, \text{K kg mol}^{-1}\)).

Calculate Molality:

Each solute has 2 moles dissolved in 500 grams of water (\(0.5 \, \text{kg}\)):

\( m_1 = \frac{2 \, \text{moles}}{0.5 \, \text{kg}} = 4 \, \text{mol kg}^{-1} \)

\( m_2 = \frac{2 \, \text{moles}}{0.5 \, \text{kg}} = 4 \, \text{mol kg}^{-1} \)

Substitute into the Formula:

\( \Delta T_b = 1 \cdot 4 \cdot 0.52 + 1 \cdot 4 \cdot 0.52 = 4.16 \)

Determine Boiling Point of Solution:

The normal boiling point of water is \(373.16 \, \text{K}\).

Add the boiling point elevation to the normal boiling point:

\( T_b (\text{solution}) = 373.16 + 4.16 = 377.3 \, \text{K} \)

Thus, the boiling point of the resulting solution is \(377.3 \, \text{K}\).

Negative Deviation from Raoult's Law

• Negative deviation means the intermolecular forces of attraction between the molecules of the two liquids (A-B) are stronger than the forces between molecules of the pure liquids (A-A and B-B).


• This stronger attraction makes it harder for molecules to escape into the vapor phase.

Effect on Vapor Pressure and Boiling Point

• Vapor Pressure : Since the molecules are held more tightly, the vapor pressure of the solution will be lower than expected from Raoult's law. Decreased vapor pressure.


• Boiling Point : A lower vapor pressure means you need to increase the temperature further to reach atmospheric pressure, where boiling occurs. Therefore, the boiling point of the solution will be higher than expected. Increased boiling point.

Answer

The correct answer is Option D: decreased vapor pressure, increased boiling point.

• Molal boiling point elevation constant of water (\(K_b\)) = 0.52 K.kg.mol^-1
• 1 atm = 760 mm Hg
• Molar mass of water = 18 g.mol^-1

First, we calculate the molality (m) of the solution using the boiling point elevation formula:

\( \Delta T_b = K_b \times m \)

From the problem, we know:

\( 2 = 0.52 \times m \)

Solving for m:

\( m = \frac{2}{0.52} \approx 3.846 \text{ mol/kg} \)

Next, considering the solution is highly diluted, we use the formula for relative lowering of vapor pressure:

\( \frac{\Delta P}{P^0} = \frac{n_{\text{solute}}}{n_{\text{solvent}}} \)

Given that molality (m) is defined as the number of moles of solute per kilogram of solvent:

\( \frac{\Delta P}{P^0} = \frac{m}{1000} \times M_{solvent} \)

Substitute the known values:

\( \Delta P = P^0 \times \frac{m}{1000} \times M_{solvent} \)

\( = 760 \times \frac{3.846}{1000} \times 18 \)

\( = 52.615 \text{ mm Hg} \)

Therefore, the vapor pressure of the solution:

\( P_{solution} = P^0 - \Delta P \)

\( = 760 - 52.615 \)

\( \approx 707.385 \text{ mm Hg} \)

Rounding to the nearest integer, the vapor pressure of the aqueous solution is 707 mm Hg.

To find the degree of ionization \((\alpha)\), we need to use the boiling point elevation formula and the van't Hoff factor. The formulas we will use are:

\(\Delta T_b = i \cdot K_b \cdot m\)

where:

\(\Delta T_b\) is the boiling point elevation,

\(i\) is the van't Hoff factor,

\(K_b\) is the molal boiling point elevation constant, and

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

Given data:

• \(\Delta T_b = 100.52^{\circ} \mathrm{C} - 100^{\circ} \mathrm{C} = 0.52^{\circ} \mathrm{C}\)
• Mass of \(\mathrm{AB}_2\) = 10 g
• Mass of water = 100 g = 0.1 kg
• Molar mass of \(\mathrm{AB}_2\) = 200 g/mol
• \(K_b\) = 0.52 K kg mol^-1

First, we calculate the molality (m):

\(m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}\)

\(= \frac{\frac{10}{200}}{0.1} \:\mathrm{mol\ kg}^{-1}\)

\(= 0.5 \:\mathrm{mol\ kg}^{-1}\)

Next, using the formula for boiling point elevation:

\(0.52 = i \cdot 0.52 \cdot 0.5\)

\(i = \frac{0.52}{0.52 \cdot 0.5}\)

\(i = \frac{1}{0.5}\)

\(i = 2\)

Now, the van't Hoff factor \(i\) is related to the degree of ionization \((\alpha)\). For the electrolyte \(\mathrm{AB}_2\), which ionizes as \(\mathrm{AB}_2 \rightarrow \mathrm{A}^{2+}+2\mathrm{B}^{-}\), the van't Hoff factor \(i\) can be expressed as:

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

where

• \(n\) is the number of ions produced (which is 3 for \(\mathrm{AB}_2 \rightarrow \mathrm{A}^{2+}+2\mathrm{B}^{-}\))
• \(\alpha\) is the degree of ionization

Substituting \(i = 2\) into the equation:

\(2 = 1 + 2 \alpha\)

\(2 - 1 = 2 \alpha\)

\(\alpha = \frac{1}{2}\)

\(\alpha = 0.5\)

The degree of ionization \((\alpha)\) is therefore

\(0.5 \times 10^{-1} = 5 \times 10^{-2}\).

Hence, the degree of ionization of the electrolyte \((\alpha)\) is approximately 5 \(\times 10^{-1}\) (the nearest integer is 5).

\(\mathrm{\Delta T_b=K_b.m}\)

(0.52) = (0.52) (m)

\(\mathrm{m=1=\frac{2(1000)}{(mw)(20)}}\)

mw = 100

Amount of solvent = 100 - (29.25 + 19) = 51.75 g

Now, we can calculate the boiling point elevation using the given formula:

ΔTb = [(2 × 29.25 × 1000) / (58.5 × 51.75) + (3 × 19 × 1000) / (95 × 51.75)] × 0.52

ΔTb = 16.075

The boiling point of the sea water is the normal boiling point of water plus the change in boiling point:

Boiling point of sea water = 100 °C + 16.075 °C = 116.075 °C

Rounding to the nearest integer, the normal boiling point of the sea water is approximately 116 °C.

The boiling point elevation constant, also known as the ebullioscopic constant (\(K_{b}\)), can be determined using the formula:

\(K_{b} = \frac{R \cdot T_{b}^{2}}{1000 \cdot \Delta H_{vap}}\)

where:

- \(R\) is the universal gas constant (8.31 J mol⁻¹ K⁻¹)

- \(T_{b}\) is the boiling point of the solvent (in K)

- \(\Delta H_{vap}\) is the enthalpy of vaporization (in J/mol)

According to the problem, the ratio of boiling points for X and Y is 2:1, and the ratio of their enthalpy of vaporization is 1:2.

Let's denote the boiling point of Y as \(T_{b}(Y)\) and the boiling point of X as \(T_{b}(X)\), and likewise for the enthalpy of vaporization \(\Delta H_{vap}(Y)\) and \(\Delta H_{vap}(X)\).

We then have:

\(T_{b}(X) = 2T_{b}(Y)\) and \(\Delta H_{vap}(X) = 0.5\Delta H_{vap}(Y)\)

We can substitute these values into the equation for \(K_{b}\):

\(K_{b}(X) = R*(2T_{b}(Y))^2 / (1000 * 0.5\Delta H_{vap}(Y)) = 4*R*T_{b}(Y)^2 / (500 * \Delta H_{vap}(Y))\)

and

\(K_{b}(Y) = R*T_{b}(Y)^2 / (1000 * \Delta H_{vap}(Y))\)

Comparing these two equations:

\(K_{b}(X) / K_{b}(Y) = [4*R*T_{b}(Y)^2 / (500 * \Delta H_{vap}(Y))] / [R*T_{b}(Y)^2 / (1000 * \Delta H_{vap}(Y))] \)

This simplifies to:

\(K_{b}(X) / K_{b}(Y) = 4 / 0.5 = 8\)

So, the boiling point elevation constant of X is 8 times the boiling point elevation constant of Y. Therefore, m = 8.

At any temperature, the vapour pressure of the solution is lower than that of the pure solvent. Hence vapour pressure- temperature curve of solution lies below that of solvent.

The more volatile liquid evaporates fast as compared to the less volatile liquid at a low temperature because the volume increases with respect to temperature so it has a low boiling point.

\(W_{\text {solute }}=2.5 \times 10^{-3} \mathrm{~kg}\)

\( \begin{aligned} &\mathrm{W}_{\text {solvent }}=75 \times 10^{-3} \mathrm{~kg} \\\\ &\Delta \mathrm{T}_{\mathrm{b}} =373.535-373.15 \\\\ &=0.385 \mathrm{~K} \\\\ &\mathrm{~K}_{\mathrm{b}}\left(\mathrm{H}_{2} \mathrm{O}\right) =0.52 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} \\\\ &\Delta \mathrm{T}_{\mathrm{b}} =\frac{\mathrm{K}_{\mathrm{b}} \times 10^{3} \times \mathrm{W}_{\text {solute }}}{\mathrm{M}_{\text {solute }} \times \mathrm{W}_{\text {solvent }}} \\\\ &\mathrm{M}_{\text {solute }} =\frac{0.52 \times 10^{3} \times 2.5 \times 10^{-3}}{75 \times 10^{-3} \times 0.385} \\\\ &=45.02 \\\\ & \approx 45 \end{aligned} \)

For \(\mathbf{A}: 100 \,\mathrm{gm}\) solution \(\rightarrow 2 \,\mathrm{gm}\) solute \(\mathrm{A}\)

\(\therefore\) Molality \(=\frac{2 / \mathrm{M}_{\mathrm{A}}}{0.098}\)

For B : \(100 \,\mathrm{gm}\) solution \(\rightarrow 8 \,\mathrm{gm}\) solute \(\mathrm{B}\)

\( \begin{aligned} &\therefore \text { Molality }=\frac{8 / \mathrm{M}_{\mathrm{B}}}{0.092} \\\\ &\because\left(\Delta \mathrm{T}_{\mathrm{B}}\right)_{\mathrm{A}}=\left(\Delta \mathrm{T}_{\mathrm{B}}\right)_{\mathrm{B}} \end{aligned} \)

\(\therefore\) Molality of \(\mathrm{A}=\) Molality of \(\mathrm{B}\)

\( \therefore \frac{2}{0.098 \mathrm{M}_{\mathrm{A}}}=\frac{8}{0.092 \mathrm{M}_{\mathrm{B}}} \)

\( \frac{2}{98} \times \frac{92}{8}=\frac{M_A}{M_B} \)

\( \frac{1}{4.261}=\frac{\mathrm{M}_{\mathrm{A}}}{\mathrm{M}_{\mathrm{B}}} \)

\( \therefore \mathrm{M}_{\mathrm{B}}=4.261 \times \mathrm{M}_{\mathrm{A}} \)