relative-lowering-of-vapour-pressure-and-roults-law

When a non-volatile solute is added to a solvent, it causes the vapour pressure of the solvent to decrease. In this scenario, when the vapour pressure decreases by 10 mm of Hg, the mole fraction of the solute in the solution is 0.2.

By understanding the relationship between vapour pressure change and mole fraction, we see that:

The change in vapour pressure (\(P^{\circ} - P\)) is directly proportional to the mole fraction of the solute (\(X_{\text{solute}}\)).

Therefore, if a 10 mm of Hg decrease corresponds to a mole fraction of 0.2, then a 20 mm of Hg decrease would correspond to a mole fraction of 0.4.

To find the mole fraction of the solvent (\(X_{\text{solvent}}\)), we use the formula:

\( X_{\text{solvent}} = 1 - X_{\text{solute}} \)

Substituting the value we found:

\( X_{\text{solvent}} = 1 - 0.4 = 0.6 \)

Thus, when the vapour pressure decreases by 20 mm of Hg, the mole fraction of the solvent is 0.6.

For a binary solution of two volatile liquid components labeled 1 and 2, let \( x_1 \) and \( y_1 \) represent the mole fractions of component 1 in the liquid and vapor phases, respectively. The linear relationship between the inverse of these mole fractions is plotted as \(\frac{1}{x_1}\) versus \(\frac{1}{y_1}\).

To derive the slope and intercept of this linear plot, consider the following calculations:

Using Raoult's Law for a Liquid Solution:

For a liquid solution with volatile components 1 and 2:

\( \mathrm{P}_1 = \mathrm{P}_{\mathrm{T}} \cdot y_1 = \mathrm{P}_1^{\mathrm{o}} \cdot x_1 \)

Therefore:

\( \frac{\mathrm{P}_{\mathrm{T}}}{x_1} = \frac{\mathrm{P}_1^{\mathrm{o}}}{y_1} \)

Rearranging the Equation:

By substituting and rearranging, we have:

\( \frac{\mathrm{P}_2^{\mathrm{o}} + x_1(\mathrm{P}_1^{\mathrm{o}} - \mathrm{P}_2^{\mathrm{o}})}{x_1} = \frac{\mathrm{P}_1^{\mathrm{o}}}{y_1} \)

Simplifying further:

\( \frac{\mathrm{P}_2^{\mathrm{o}}}{x_1} + (\mathrm{P}_1^{\mathrm{o}} - \mathrm{P}_2^{\mathrm{o}}) = \frac{\mathrm{P}_1^{\mathrm{o}}}{y_1} \)

Expressing \(\frac{1}{x_1}\):

Solving for \(\frac{1}{x_1}\), we obtain:

\( \frac{1}{x_1} = \left(\frac{\mathrm{P}_1^{\mathrm{o}}}{\mathrm{P}_2^{\mathrm{o}}}\right)\left(\frac{1}{y_1}\right) + \left(\frac{\mathrm{P}_2^{\mathrm{o}} - \mathrm{P}_1^{\mathrm{o}}}{\mathrm{P}_2^{\mathrm{o}}}\right) \)

Determining the Slope and Intercept:

The slope of the line is:

\( \text{Slope} = \frac{\mathrm{P}_1^{\mathrm{o}}}{\mathrm{P}_2^{\mathrm{o}}} \)

The intercept of the line is:

\( \text{Intercept} = \frac{\mathrm{P}_2^{\mathrm{o}} - \mathrm{P}_1^{\mathrm{o}}}{\mathrm{P}_2^{\mathrm{o}}} \)

In summary, for the plot of \(\frac{1}{x_1}\) against \(\frac{1}{y_1}\), the slope is \(\frac{\mathrm{P}_1^{\mathrm{o}}}{\mathrm{P}_2^{\mathrm{o}}}\) and the intercept is \(\frac{\mathrm{P}_2^{\mathrm{o}} - \mathrm{P}_1^{\mathrm{o}}}{\mathrm{P}_2^{\mathrm{o}}}\).

A binary mixture of \( \text{C}_6\text{H}_5\text{OH} \) and \( \text{C}_6\text{H}_5\text{NH}_2 \) exhibits negative deviation from Raoult's law. This means that the vapor pressure of the solution is lower than the vapor pressures of the pure components, \( \text{C}_6\text{H}_5\text{OH} \) and \( \text{C}_6\text{H}_5\text{NH}_2 \). Consequently, the boiling point of the solution is higher than the boiling points of the pure substances. Therefore, this mixture forms a maximum boiling azeotrope.

When \(x\) grams of NaCl are added to water in a beaker and the temperature is increased from \(1^{\circ} \mathrm{C}\) to \(25^{\circ} \mathrm{C}\), the molarity of the solution changes due to the volumetric changes of water.

Temperature from \(1^{\circ} \mathrm{C}\) to \(4^{\circ} \mathrm{C}\):

Water is densest at \(4^{\circ} \mathrm{C}\).

As the temperature increases from \(1^{\circ} \mathrm{C}\) to \(4^{\circ} \mathrm{C}\), the water volume decreases due to increased density.

This decrease in volume results in an increase in molarity because molarity is inversely proportional to the solution's volume.

Temperature from \(4^{\circ} \mathrm{C}\) to \(25^{\circ} \mathrm{C}\):

Beyond \(4^{\circ} \mathrm{C}\), water expands with an increase in temperature.

Therefore, as temperature rises to \(25^{\circ} \mathrm{C}\), the volume of the water increases.

The dilution leads to a decrease in molarity, since molarity is inversely proportional to volume.

Thus, the molarity first increases as temperature rises to \(4^{\circ} \mathrm{C}\), but then decreases as it continues to \(25^{\circ} \mathrm{C}\). The graphical representation of this relationship would exhibit an initial increase in molarity, followed by a decrease, correlating with changes in the volume of water due to temperature variations.

To correctly match the items from List - I with those in List - II, let's analyze the characteristics of each solution:

(A) Solution of chloroform and acetone: This solution exhibits negative deviation from Raoult's law. It creates a maximum boiling azeotrope because the interactions between chloroform and acetone molecules are stronger than those in the pure components.

(B) Solution of ethanol and water: This solution shows positive deviation from Raoult's law, leading to the formation of a minimum boiling azeotrope. The interactions between ethanol and water molecules are weaker than those in their pure states.

(C) Solution of benzene and toluene: This combination forms an ideal solution where Raoult’s law is obeyed across all concentrations. Thus, the volume change upon mixing, denoted as \(\Delta V_{\text{mix}}\), is zero.

(D) Solution of acetic acid in benzene: In this mixture, acetic acid tends to dimerize. The acetic acid molecules pair up, forming dimers, especially in non-polar solvents like benzene.

Based on these explanations, the correct matches are:

(A) - (III)

(B) - (I)

(C) - (IV)

(D) - (II)

\(\left(\mathrm{CH}_3\right)_2 \mathrm{CO}+\mathrm{CS}_2\) Exibits positive deviations from Raoult's Law

To find the mole fraction of methyl benzene in the vapor phase when it is in equilibrium with an equimolar (equal mole) mixture of benzene and methyl benzene at \(27^{\circ}C\), we can apply Raoult's law for an ideal solution. Given the vapor pressures of pure benzene and methyl benzene are 80 Torr and 24 Torr, respectively.

Raoult's law states that the partial vapor pressure of a component in a solution is equal to the vapor pressure of the pure component multiplied by its mole fraction in the solution (\(P_i = P_i^\circ x_i\)), where \(P_i\) is the partial vapor pressure of the component \(i\), \(P_i^\circ\) is the vapor pressure of the pure component \(i\), and \(x_i\) is the mole fraction of the component \(i\) in the solution.

For an equimolar mixture of benzene and methyl benzene, the mole fractions of benzene (\(x_b\)) and methyl benzene (\(x_{mb}\)) in the liquid phase are both \(0.5\), because the mixture is equimolar.

The total pressure of the mixture (\(P_{total}\)) can be calculated using Raoult's law for each component and summing up the individual pressures:

\(P_{total} = P_{benzene} + P_{methyl\ benzene}\)

\(P_{total} = (P_b^\circ x_b) + (P_{mb}^\circ x_{mb})\)

Substituting the given values:

\(P_{total} = (80 \times 0.5) + (24 \times 0.5)\)

\(P_{total} = 40 + 12 = 52 \, \text{Torr}\)

The mole fraction of methyl benzene in the vapor phase (\(y_{mb}\)) can be calculated using Dalton's law, which states that the partial pressure of a component in a mixture is equal to the mole fraction of that component in the vapor phase times the total pressure of the mixture.

The partial pressure of methyl benzene can be obtained from Raoult's law as we did earlier:

\(P_{methyl\ benzene} = P_{mb}^\circ x_{mb} = 24 \times 0.5 = 12 \, \text{Torr}\)

Using Dalton's law:

\(y_{mb} = \frac{P_{methyl\ benzene}}{P_{total}}\)

Substituting the values:

\(y_{mb} = \frac{12}{52}\)

\(y_{mb} = \frac{6}{26}\)

\(y_{mb} = \frac{3}{13}\)

To express \(y_{mb}\) as a percentage times \(10^{-2}\):

\(y_{mb} = \left(\frac{3}{13}\right) \times 100 \times 10^{-2}\)

\(y_{mb} \approx 23.08 \times 10^{-2}\)

So, the mole fraction of methyl benzene in the vapor phase, in equilibrium with an equimolar mixture of those two liquids at \(27^{\circ}C\), is approximately \(23 \times 10^{-2}\) (rounded to the nearest integer).

The given solution is a \(30 \%\) (w/v) aqueous solution of glucose. This means that \(30\ \mathrm{g}\) of glucose is dissolved in \(100\ \mathrm{mL}\) of the solution. Since the density of the solution is \(1.2\ \mathrm{g/mL}\), the weight of \(100\ \mathrm{mL}\) of the solution is:

\(\mathrm{Wt.~of~solution} = 100\ \mathrm{mL} \times 1.2\ \mathrm{g/mL} = 120\ \mathrm{g}\)

Since the solution is \(30 \%\) (w/v), the weight of glucose in \(100\ \mathrm{mL}\) of the solution is \(30\ \mathrm{g}\). Therefore, the weight of water in \(100\ \mathrm{mL}\) of the solution is:

\(\mathrm{Wt.~of~water} = 120\ \mathrm{g} - 30\ \mathrm{g} = 90\ \mathrm{g}\)

Now, we can use Raoult's law to calculate the vapour pressure of the solution. Raoult's law states that the partial vapour pressure of a solvent in a solution is proportional to its mole fraction in the solution. For a dilute solution, the mole fraction of the solvent can be approximated as the ratio of the number of moles of solvent to the total number of moles in the solution.

Let us assume that we have 1 mole of the solution. The number of moles of glucose in the solution is:

\(n_{\mathrm{glucose}} = \frac{30\ \mathrm{g}}{180\ \mathrm{g/mol}} = 0.167\ \mathrm{mol}\)

The number of moles of water in the solution is:

\(n_{\mathrm{water}} = \frac{90\ \mathrm{g}}{18\ \mathrm{g/mol}} = 5\ \mathrm{mol}\)

The total number of moles in the solution is:

\(n_{\mathrm{total}} = n_{\mathrm{glucose}} + n_{\mathrm{water}} = 5.167\ \mathrm{mol}\)

Let \(P\) be the vapour pressure of the solution. According to Raoult's law, we have:

\(\frac{P_0 - P}{P} = \frac{n_{\mathrm{glucose}}}{n_{\mathrm{water}}} = \frac{0.167}{5} = 0.0334\)

where \(P_0\) is the vapour pressure of pure water, which is given as \(24\ \mathrm{mmHg}\).

Simplifying the above equation, we get:

\(P = \frac{P_0}{1 + \frac{n_{\mathrm{glucose}}}{n_{\mathrm{water}}}} = \frac{24\ \mathrm{mmHg}}{1 + \frac{0.167}{5}} = 23.22\ \mathrm{mmHg}\)

Therefore, the vapour pressure of the \(30\%\) (w/v) aqueous solution of glucose at \(25^\circ\mathrm{C}\) is \(\boxed{23\ \mathrm{mmHg}}\) (rounded off to the nearest integer).

\( \begin{array}{ccc} \mathrm{MgCl}_2 & \rightleftharpoons ~\mathrm{Mg}^{2+}+2 \mathrm{Cl}^{-} \\ 1 & 0 & 0 \\ 1-0.8 & 0.8 & 1.6 \end{array} \)

Hence overall molality will be equal to \(=2.6\)

\( \frac{p^{\circ}-p}{p^{\circ}}=\frac{2.6}{\frac{1000}{18}+2.6} \)

For dil solution

\( \begin{aligned} & \frac{p^{\circ}-p}{p^{\circ}}=\frac{2.6}{\frac{1000}{18}} \\\\ & p=47.66 \simeq 48 \mathrm{~mm} \mathrm{Hg} \end{aligned} \)

Given:

• Vapor pressure reduction: \(25\%\) (\(0.75\) times the vapor pressure of pure water)


• Molar mass of water (\(\text{H}_2\text{O}\)): \(18 \, \text{g/mol}\)


• Mass of solvent (water): \(1000 \, \text{g}\)

Using Raoult's law:

\(\frac{P^0 - P_s}{P_s} = \frac{n_{\text{solute}}}{n_{\text{solvent}}} = \frac{\frac{x}{M_{\text{urea}}}}{\frac{1000}{M_{\text{water}}}} = \frac{P^0 - 0.75P^0}{0.75P^0}\)

Solving for (x):

\(\frac{x}{60} = \frac{10000}{9}\) \(x = 1111.11111 \approx 1111 \, \text{g}\)

So, the mass of urea required to be dissolved in \(1000 \, \text{g}\) of water is \(1111 \, \text{g}\).