Lowering of vapour pressure Questions in English

(N/A) Given: $P_A^0 = 450 \ mm \ Hg$,$P_B^0 = 700 \ mm \ Hg$,$P_{total} = 600 \ mm \ Hg$.
Let the mole fraction of $A$ in the liquid phase be $x_A$. Then the mole fraction of $B$ is $x_B = 1 - x_A$.
According to Raoult's Law: $P_{total} = P_A^0 x_A + P_B^0 x_B$.
$600 = 450 x_A + 700(1 - x_A)$.
$600 = 450 x_A + 700 - 700 x_A$.
$250 x_A = 100$.
$x_A = 100 / 250 = 0.4$.
So,$x_B = 1 - 0.4 = 0.6$.
The partial pressures are: $P_A = P_A^0 x_A = 450 \times 0.4 = 180 \ mm \ Hg$ and $P_B = P_B^0 x_B = 700 \times 0.6 = 420 \ mm \ Hg$.
In the vapour phase,the mole fractions are $y_A = P_A / P_{total} = 180 / 600 = 0.3$ and $y_B = P_B / P_{total} = 420 / 600 = 0.7$.

(N/A) Raoult's law states that for a solution of volatile liquids,the partial vapour pressure of each component in the solution is directly proportional to its mole fraction present in the solution.
For a solution containing a non-volatile solute,only the solvent contributes to the vapour pressure. The law is expressed as: $p_{1} = x_{1} \cdot p_{1}^{0}$,where $p_{1}$ is the vapour pressure of the solvent in the solution,$x_{1}$ is its mole fraction,and $p_{1}^{0}$ is the vapour pressure of the pure solvent.
The reduction in vapour pressure $(\Delta p_{1})$ is given by: $\Delta p_{1} = p_{1}^{0} - p_{1} = p_{1}^{0} - p_{1}^{0} x_{1} = p_{1}^{0}(1 - x_{1})$.
Since $1 - x_{1} = x_{2}$ (where $x_{2}$ is the mole fraction of the solute),we have $\Delta p_{1} = x_{2} \cdot p_{1}^{0}$.
The relative lowering of vapour pressure is given by: $\frac{p_{1}^{0} - p_{1}}{p_{1}^{0}} = x_{2} = \frac{n_{2}}{n_{1} + n_{2}}$.
For dilute solutions,$n_{2} \ll n_{1}$,so $\frac{p_{1}^{0} - p_{1}}{p_{1}^{0}} \approx \frac{n_{2}}{n_{1}} = \frac{w_{2} \times M_{1}}{M_{2} \times w_{1}}$,where $w$ is mass and $M$ is molar mass.

(N/A) The vapour pressure of a liquid is determined by the escaping tendency of its molecules from the surface. In pure water,the entire surface is occupied by water molecules. When a non-volatile solute like glucose is added,the glucose molecules occupy some space on the surface. This reduces the number of water molecules available at the surface to escape into the vapour phase. Consequently,the rate of evaporation decreases,leading to a lower vapour pressure for the solution compared to pure water. This phenomenon is a colligative property.

(N/A) For a binary solution where both components are volatile liquids ($CHCl_{3}$ and $CH_{2}Cl_{2}$),the total vapour pressure $(p_{total})$ is given by the sum of the partial vapour pressures of the components:
$p_{total} = p_{1} + p_{2} = x_{1}p_{1}^{0} + x_{2}p_{2}^{0}$
Since $x_{1} + x_{2} = 1$,we can write $p_{total} = x_{1}p_{1}^{0} + (1 - x_{1})p_{2}^{0} = x_{1}(p_{1}^{0} - p_{2}^{0}) + p_{2}^{0}$.
Thus,the total vapour pressure varies linearly with the mole fraction of one component.
$(b)$ For a solution containing a non-volatile solute $(NaCl_{(s)})$ and a volatile solvent $(H_{2}O_{(l)})$,the solute does not contribute to the vapour pressure.
According to Raoult's law,the total vapour pressure of the solution is equal to the partial vapour pressure of the solvent:
$p_{total} = p_{solvent} = x_{solvent}p_{solvent}^{0}$
Here,the total vapour pressure is directly proportional to the mole fraction of the solvent $(H_{2}O_{(l)})$.

(N/A) $(i)$ Boiling point is the temperature at which vapour pressure equals atmospheric pressure $(760 \ mm \ Hg)$. From the graph,for liquid $A$,the boiling point is approximately $315 \ K$,and for liquid $B$,it is approximately $345 \ K$.
(ii) Liquid $C$ will not boil in a closed vessel because as the temperature increases,the vapour pressure increases,but the pressure inside the vessel also increases simultaneously,preventing the liquid from reaching its boiling point.
(iii) At $60 \ mm \ Hg$,by observing the graph for liquid $D$,the corresponding temperature is approximately $313 \ K$.
(iv) $A$ liquid boils when its vapour pressure equals the external atmospheric pressure. At high altitudes,atmospheric pressure is low,so water boils at a lower temperature,which is insufficient for cooking food properly. $A$ pressure cooker increases the internal pressure,which in turn increases the boiling point of water,allowing food to cook faster at a higher temperature.

(N/A) Vapor pressure is directly proportional to temperature $(T)$.
At $373 \, K$ $(100^{\circ} \, C)$,which is the boiling point of water at standard atmospheric pressure,the vapor pressure is equal to $1 \, atm$.

(A) Vapor pressure depends on the strength of intermolecular forces.
Ether has weak dipole-dipole interactions,resulting in the highest vapor pressure.
Ethanol exhibits strong hydrogen bonding,which restricts the escape of molecules into the vapor phase,resulting in the lowest vapor pressure.
Acetone has intermediate dipole-dipole interactions.
Therefore,the order of vapor pressure is: $\text{Ether} > \text{Acetone} > \text{Ethanol}$.

(C) There will be no change in the vapor pressure.
Vapor pressure is an intensive property that depends only on the temperature and the nature of the liquid.
Since the temperature remains constant at $300 \ K$ and the intermolecular forces within the liquid do not change,the vapor pressure remains constant regardless of the size of the container.

(C) When a non-volatile solute like glucose is added to a solvent,the number of solvent molecules at the surface decreases.
This results in a decrease in the rate of evaporation,meaning fewer water molecules leave the solution.
Consequently,the vapour pressure of the solution decreases.

(A) Let $P_{A}^{\circ}$ be the vapour pressure of pure $n-hexane$ and $P_{B}^{\circ}$ be the vapour pressure of pure $n-heptane$.
According to Raoult's Law,$P_{total} = X_{A}P_{A}^{\circ} + X_{B}P_{B}^{\circ}$.
For the first case: $X_{A} = \frac{1}{1+3} = \frac{1}{4}$ and $X_{B} = \frac{3}{4}$.
$550 = \frac{1}{4}P_{A}^{\circ} + \frac{3}{4}P_{B}^{\circ} \implies P_{A}^{\circ} + 3P_{B}^{\circ} = 2200 \dots (i)$
For the second case: $1 \ mole$ of $n-heptane$ is added,so total moles = $1 + 4 = 5$. $X_{A} = \frac{1}{5}$ and $X_{B} = \frac{4}{5}$.
The new vapour pressure is $550 + 10 = 560 \ mm \ Hg$.
$560 = \frac{1}{5}P_{A}^{\circ} + \frac{4}{5}P_{B}^{\circ} \implies P_{A}^{\circ} + 4P_{B}^{\circ} = 2800 \dots (ii)$
Subtracting equation $(i)$ from $(ii)$:
$(P_{A}^{\circ} + 4P_{B}^{\circ}) - (P_{A}^{\circ} + 3P_{B}^{\circ}) = 2800 - 2200$
$P_{B}^{\circ} = 600 \ mm \ Hg$.

(A) According to Raoult's law,the relative lowering of vapour pressure is equal to the mole fraction of the solute: $\frac{\Delta P}{P^0} = x_{\text{solute}} = \frac{n_{\text{solute}}}{n_{\text{solute}} + n_{\text{solvent}}}$.
Since the mass of the solvent ($180 \ g$ of water) and the mass of the solutes $(10 \ g)$ are constant,the mole fraction of the solute depends on the number of moles of the solute $(n = \frac{\text{mass}}{\text{molar mass}})$.
As the molar mass of the solute increases,the number of moles of the solute decreases,which in turn decreases the mole fraction of the solute.
Given molar masses: $M_A = 100 \ g \ mol^{-1}$,$M_B = 200 \ g \ mol^{-1}$,$M_C = 10,000 \ g \ mol^{-1}$.
Since $M_A < M_B < M_C$,the number of moles follows $n_A > n_B > n_C$.
Therefore,the relative lowering of vapour pressure follows the order $A > B > C$.

(B) According to Raoult's law for non-volatile solutes,the relative lowering of vapour pressure is given by:
$\frac{P_{A}^{0} - P_{s}}{P_{A}^{0}} = \frac{n_{B}}{n_{A} + n_{B}} \approx \frac{n_{B}}{n_{A}}$ (for dilute solutions).
Given:
$w_{B} = 8 \; g$,$W_{A} = 114 \; g$,$M_{A} = 114 \; g \; mol^{-1}$.
Since the vapour pressure is reduced to $80 \%$,$P_{s} = 0.8 P_{A}^{0}$.
Therefore,$\frac{P_{A}^{0} - 0.8 P_{A}^{0}}{P_{A}^{0}} = \frac{0.2 P_{A}^{0}}{P_{A}^{0}} = 0.2$.
Using the formula $\frac{\Delta P}{P_{A}^{0}} = \frac{w_{B} \times M_{A}}{m_{B} \times W_{A}}$:
$0.2 = \frac{8 \times 114}{m_{B} \times 114}$.
$0.2 = \frac{8}{m_{B}}$.
$m_{B} = \frac{8}{0.2} = 40 \; g \; mol^{-1}$.

(C) The molar mass of $CHCl_3$ is $119.5 \,g/mol$ and the molar mass of $CH_2Cl_2$ is $85 \,g/mol$.
Number of moles of $CHCl_3$ $(n_1)$ = $\frac{11.9 \,g}{119.5 \,g/mol} \approx 0.1 \,mol$.
Number of moles of $CH_2Cl_2$ $(n_2)$ = $\frac{17 \,g}{85 \,g/mol} = 0.2 \,mol$.
Total moles = $0.1 + 0.2 = 0.3 \,mol$.
Mole fraction of $CHCl_3$ $(x_1)$ = $\frac{0.1}{0.3} = \frac{1}{3}$.
Mole fraction of $CH_2Cl_2$ $(x_2)$ = $\frac{0.2}{0.3} = \frac{2}{3}$.
According to Raoult's law,the total vapour pressure $(P_T)$ is:
$P_T = P_1^{\circ} x_1 + P_2^{\circ} x_2$
$P_T = (200 \times \frac{1}{3}) + (41.5 \times \frac{2}{3})$
$P_T = 66.67 + 27.67 = 94.34 \,atm \approx 94.3 \,atm$.

(A) The relative lowering of vapour pressure is given by the formula: $\frac{p^{\circ} - p_{s}}{p^{\circ}} = x_{2} = \frac{n_{2}}{n_{1} + n_{2}} \approx \frac{n_{2}}{n_{1}}$.
First,calculate the moles of solute $(n_{2})$: $n_{2} = \frac{0.5}{65} \approx 0.00769 \, mol$.
Next,calculate the mass of $CCl_{4}$: $Mass = Density \times Volume = 1.58 \, g/cm^{3} \times 100 \, cm^{3} = 158 \, g$.
Molar mass of $CCl_{4} = 12 + 4 \times 35.5 = 154 \, g/mol$.
Moles of solvent $(n_{1})$: $n_{1} = \frac{158}{154} \approx 1.026 \, mol$.
Using the formula: $\frac{143 - p_{s}}{143} = \frac{0.00769}{1.026}$.
$143 - p_{s} = 143 \times 0.007495 = 1.0718$.
$p_{s} = 143 - 1.0718 = 141.9282 \, mm\, Hg \approx 141.93 \, mm\, Hg$.

(B) Step $1$: Calculate moles of reactants.
$n_{SO_2} = \frac{224 \, mL}{22400 \, mL/mol} = 0.01 \, mol$.
$n_{NaOH} = 0.1 \, M \times 0.1 \, L = 0.01 \, mol$.
Step $2$: Reaction stoichiometry.
$SO_2 + NaOH \rightarrow NaHSO_3$.
Since $n_{SO_2} = n_{NaOH} = 0.01 \, mol$, both are consumed to produce $0.01 \, mol$ of $NaHSO_3$.
Step $3$: Calculate lowering of vapour pressure.
$\Delta P = P^0_{H_2O} \times \chi_{solute} = P^0_{H_2O} \times \frac{n_{solute}}{n_{solute} + n_{solvent}}$.
$n_{solvent} = \frac{36 \, g}{18 \, g/mol} = 2 \, mol$.
$\Delta P = 24 \times \frac{0.01}{0.01 + 2} = 24 \times \frac{0.01}{2.01} \approx 24 \times 0.004975 = 0.1194 \, mm \, Hg$.
$\Delta P = 11.94 \times 10^{-2} \, mm \, Hg$.
Rounding to the nearest integer, $x = 12$.

(C) According to Raoult's law for an ideal solution,the total vapour pressure $P_s$ is given by:
$P_s = P_A^0 x_A + P_B^0 x_B$
Given:
$P_A^0$ (benzene) $= 280 \ mm \ Hg$
$P_B^0$ (octane) $= 420 \ mm \ Hg$
Molar ratio $= 3 : 2$,so mole fractions are:
$x_A = \frac{3}{3+2} = \frac{3}{5} = 0.6$
$x_B = \frac{2}{3+2} = \frac{2}{5} = 0.4$
Substituting the values:
$P_s = (280 \times 0.6) + (420 \times 0.4)$
$P_s = 168 + 168$
$P_s = 336 \ mm \ Hg$

(D) Given: $P_{B}^{\circ} = 70 \, torr$,$P_{M}^{\circ} = 20 \, torr$.
Since the mixture is equimolar,the mole fractions in the liquid phase are $X_{B} = 0.5$ and $X_{M} = 0.5$.
The partial pressure of benzene is $P_{B} = X_{B} \times P_{B}^{\circ} = 0.5 \times 70 = 35 \, torr$.
The partial pressure of methyl benzene is $P_{M} = X_{M} \times P_{M}^{\circ} = 0.5 \times 20 = 10 \, torr$.
The total pressure is $P_{total} = P_{B} + P_{M} = 35 + 10 = 45 \, torr$.
The mole fraction of benzene in the vapor phase $(y_{B})$ is given by $y_{B} = \frac{P_{B}}{P_{total}} = \frac{35}{45} = 0.777...$.
Rounding to the nearest integer for $y_{B} \times 10^{-2}$,we get $78 \times 10^{-2}$.

(A) Given: $P_{A}^{\circ} = 90 \ mm \ Hg$,$P_{B}^{\circ} = 15 \ mm \ Hg$ at $25^{\circ} C$.
Mole fraction of $A$ in liquid phase,$X_{A} = 0.6$.
Mole fraction of $B$ in liquid phase,$X_{B} = 1 - 0.6 = 0.4$.
Total vapour pressure,$P_{T} = X_{A} P_{A}^{\circ} + X_{B} P_{B}^{\circ}$.
$P_{T} = (0.6 \times 90) + (0.4 \times 15) = 54 + 6 = 60 \ mm \ Hg$.
Partial pressure of $B$ in vapour phase,$P_{B} = X_{B} P_{B}^{\circ} = 0.4 \times 15 = 6 \ mm \ Hg$.
Mole fraction of $B$ in vapour phase,$Y_{B} = \frac{P_{B}}{P_{T}} = \frac{6}{60} = 0.1$.
Given $Y_{B} = x \times 10^{-1}$,so $0.1 = x \times 10^{-1}$,which gives $x = 1$.

(B) Let $A$ be toluene and $B$ be benzene.
Given: $X_{A} = 0.50$,$X_{B} = 1 - 0.50 = 0.50$.
$P_{A}^{0} = 37.0 \ torr$,$P_{B}^{0} = 119 \ torr$.
Total vapour pressure $P_{total} = P_{A}^{0} X_{A} + P_{B}^{0} X_{B} = (37.0 \times 0.50) + (119 \times 0.50) = 18.5 + 59.5 = 78.0 \ torr$.
According to Dalton's Law,the mole fraction of toluene in the vapour phase $(y_{A})$ is given by $y_{A} = \frac{P_{A}}{P_{total}}$.
$P_{A} = P_{A}^{0} X_{A} = 37.0 \times 0.50 = 18.5 \ torr$.
$y_{A} = \frac{18.5}{78.0} \approx 0.237$.

(C) Given: $P_A^{\circ} = 50 \ Torr$,$P_B^{\circ} = 100 \ Torr$,$X_A = 0.3$,$X_B = 1 - 0.3 = 0.7$.
According to Raoult's Law,the partial pressures are:
$P_A = P_A^{\circ} X_A = 50 \times 0.3 = 15 \ Torr$.
$P_B = P_B^{\circ} X_B = 100 \times 0.7 = 70 \ Torr$.
Total pressure $P_{total} = P_A + P_B = 15 + 70 = 85 \ Torr$.
The mole fraction of $B$ in the vapour phase $(y_B)$ is given by:
$y_B = \frac{P_B}{P_{total}} = \frac{70}{85} = \frac{14}{17}$.
Comparing this with $\frac{x}{17}$,we get $x = 14$.

(D) According to Raoult's law for non-volatile solutes,the relative lowering of vapour pressure is given by: $\frac{P^{0} - P_{s}}{P^{0}} = \frac{n_{A}}{n_{A} + n_{solvent}}$.
Given that the vapour pressure of the solution $P_{s}$ is half of the pure water $P^{0}$,so $P_{s} = \frac{P^{0}}{2}$.
Substituting the values: $\frac{P^{0} - P^{0}/2}{P^{0}} = \frac{n_{A}}{n_{A} + n_{solvent}} \implies \frac{1}{2} = \frac{n_{A}}{n_{A} + n_{solvent}}$.
This simplifies to $n_{A} + n_{solvent} = 2n_{A}$,which means $n_{A} = n_{solvent}$.
The number of moles of water $(n_{solvent})$ is $\frac{100 \ g}{18 \ g/mol} = 5.55 \ mol$.
Therefore,$n_{A} = 5.55 \ mol$.
Rounding to the nearest integer,we get $6$.

(C) According to Dalton's Law of partial pressures,the partial pressure of $A$ is given by $P_A = Y_A \times P_{total}$.
Given $Y_A = 0.5$ and $P_{total} = 0.8 \ atm$,we have $P_A = 0.5 \times 0.8 = 0.4 \ atm$.
According to Raoult's Law,$P_A = P_A^0 \times X_A$,where $X_A$ is the mole fraction in the liquid phase.
Given $X_A = 0.2$,we have $0.4 = P_A^0 \times 0.2$.
Therefore,$P_A^0 = \frac{0.4}{0.2} = 2 \ atm$.

(B) According to Raoult's law,for an ideal solution of liquids $X$ and $Y$:
$p_X = p_X^0 \chi_X$
$p_Y = p_Y^0 \chi_Y = p_Y^0 (1 - \chi_X)$
According to Dalton's law of partial pressures,the total vapour pressure $p_T$ is:
$p_T = p_X + p_Y$
$p_T = p_X^0 \chi_X + p_Y^0 (1 - \chi_X)$
$p_T = p_Y^0 + (p_X^0 - p_Y^0) \chi_X$
This equation represents a linear relationship between $p_T$ and $\chi_X$,which is of the form $y = mx + c$,where the slope is $(p_X^0 - p_Y^0)$ and the intercept is $p_Y^0$.
Thus,the plot of total vapour pressure versus mole fraction of $X$ is a straight line.

(D) The relative lowering of vapour pressure is given by the formula: $\frac{p_0 - p_s}{p_0} = \chi_2 = \frac{n_2}{n_1 + n_2}$.
Given: $p_0 = 760 \, mm \, Hg$ (at $100^{\circ} C$),$p_s = 750 \, mm \, Hg$,$w_2 = 18 \, g$,$w_1 = 108 \, g$,$M_1 = 18 \, g \, mol^{-1}$.
$n_1 = \frac{108}{18} = 6 \, mol$.
$n_2 = \frac{18}{M_2}$.
Using the formula $\frac{760 - 750}{760} = \frac{n_2}{n_1 + n_2} \approx \frac{n_2}{n_1}$ (since $n_2$ is small compared to $n_1$):
$\frac{10}{760} = \frac{18/M_2}{6}$.
$\frac{1}{76} = \frac{3}{M_2}$.
$M_2 = 3 \times 76 = 228 \, g \, mol^{-1}$.

(D) According to Raoult's law,the total vapour pressure $p_T$ is given by:
$p_T = x_A p_A^{\circ} + x_B p_B^{\circ}$
Since $x_B = 1 - x_A$,we have:
$p_T = x_A p_A^{\circ} + (1 - x_A) p_B^{\circ}$
Substituting the given values:
$300 = x_A(100) + (1 - x_A)(500)$
$300 = 100x_A + 500 - 500x_A$
$400x_A = 200$
$x_A = 1 / 2$
Thus,the mole fraction of $A$ in the liquid phase is $1 / 2$.
In the vapour phase,the mole fraction of $A$ $(y_A)$ is given by:
$y_A = \frac{p_A}{p_T} = \frac{x_A p_A^{\circ}}{p_T}$
$y_A = \frac{(1 / 2) \times 100}{300} = \frac{50}{300} = 1 / 6$
Therefore,the mole fractions of $A$ in the liquid and vapour phases are $1 / 2$ and $1 / 6$ respectively.

(B) The molar mass of ethylene glycol $(HOCH_2CH_2OH)$ is $62 \ g/mol$.
Number of moles of ethylene glycol $(n_{solute}) = \frac{651 \ g}{62 \ g/mol} = 10.5 \ mol$.
Number of moles of water $(n_{solvent}) = \frac{1500 \ g}{18 \ g/mol} \approx 83.33 \ mol$.
Using Raoult's law for a non-volatile solute: $\frac{P^{\circ} - P}{P^{\circ}} = \frac{n_{solute}}{n_{solute} + n_{solvent}}$.
$\frac{0.7 - P}{0.7} = \frac{10.5}{10.5 + 83.33} = \frac{10.5}{93.83} \approx 0.1119$.
$0.7 - P = 0.7 \times 0.1119 \approx 0.0783$.
$P = 0.7 - 0.0783 = 0.6217 \ atm$.
Thus,the vapour pressure is closest to $0.62 \ atm$.

(D) The relative lowering of vapour pressure is given by the formula: $\frac{P^0 - P_s}{P^0} = \frac{n_2}{n_1 + n_2} \approx \frac{n_2}{n_1}$ for dilute solutions.
Given that the vapour pressure is reduced by $25 \ \%$,the final vapour pressure $P_s = 0.75 P^0$.
Thus,$\frac{P^0 - 0.75 P^0}{P^0} = \frac{n_2}{n_1} = 0.25$.
Here,$n_2 = \frac{x}{60}$ (moles of urea) and $n_1 = \frac{1000}{18} \approx 55.56$ (moles of water).
$\frac{x/60}{1000/18} = 0.25 \Rightarrow \frac{x}{60} = 0.25 \times 55.56 = 13.89$.
$x = 13.89 \times 60 = 833.4 \ g$.
Wait,re-evaluating the formula: $\frac{P^0 - P_s}{P_s} = \frac{n_2}{n_1} = \frac{0.25 P^0}{0.75 P^0} = \frac{1}{3}$.
$\frac{x/60}{1000/18} = \frac{1}{3} \Rightarrow \frac{x}{60} = \frac{1}{3} \times 55.56 = 18.52$.
$x = 18.52 \times 60 = 1111.2 \ g$.
Rounding to the nearest integer,we get $1111 \ g$.

(D) For a dilute solution,the relative lowering of vapour pressure is given by: $\frac{P^0 - P_s}{P^0} = \frac{n}{N}$
Where $P^0 - P_s = 0.20 \ mm \ Hg$,$P^0 = 54.2 \ mm \ Hg$,$n$ is the moles of glucose,and $N$ is the moles of water.
$N = \frac{100 \ g}{18 \ g \ mol^{-1}} = 5.55 \ mol$.
Substituting the values: $\frac{0.20}{54.2} = \frac{w / 180}{100 / 18}$.
$\frac{0.20}{54.2} = \frac{w \times 18}{180 \times 100}$.
$w = \frac{0.20 \times 180 \times 100}{54.2 \times 18} = \frac{3600}{975.6} \approx 3.69 \ g$.

(A) The relative lowering of vapour pressure is given by $\frac{P^{\circ} - P_{s}}{P_{s}} = \frac{n_{solute}}{n_{solvent}}$.
Given: $30 \% \ (w/v)$ solution means $30 \ g$ of glucose in $100 \ mL$ of solution.
Mass of solution $= \text{density} \times \text{volume} = 1.2 \ g \ cm^{-3} \times 100 \ mL = 120 \ g$.
Mass of solvent (water) $= 120 \ g - 30 \ g = 90 \ g$.
Moles of glucose $(n_{solute})$ $= \frac{30 \ g}{180 \ g \ mol^{-1}} = \frac{1}{6} \ mol \approx 0.1667 \ mol$.
Moles of water $(n_{solvent})$ $= \frac{90 \ g}{18 \ g \ mol^{-1}} = 5 \ mol$.
Using the formula $\frac{24 - P_{s}}{P_{s}} = \frac{0.1667}{5} = 0.03334$.
$24 - P_{s} = 0.03334 \ P_{s} \implies 24 = 1.03334 \ P_{s}$.
$P_{s} = \frac{24}{1.03334} \approx 23.22 \ mm \ Hg$. The closest integer value is $23 \ mm \ Hg$.

(C) Given that the mixture is equimolar,the mole fraction of benzene $(X_B)$ and methyl benzene $(X_M)$ in the liquid phase are both $0.5$.
The partial vapour pressure of methyl benzene is $P_M = X_M \times P^{\circ}_M = 0.5 \times 24 \ Torr = 12 \ Torr$.
The partial vapour pressure of benzene is $P_B = X_B \times P^{\circ}_B = 0.5 \times 80 \ Torr = 40 \ Torr$.
The total vapour pressure is $P_{\text{total}} = P_M + P_B = 12 \ Torr + 40 \ Torr = 52 \ Torr$.
The mole fraction of methyl benzene in the vapour phase $(Y_M)$ is given by $Y_M = \frac{P_M}{P_{\text{total}}}$.
$Y_M = \frac{12}{52} \approx 0.2307$.
Rounding to the nearest integer as requested,$Y_M \approx 23 \times 10^{-2}$.

(A) According to Raoult's law,the relative lowering of vapour pressure is equal to the mole fraction of the solute: $\frac{P^{\circ} - P}{P^{\circ}} = X_{\text{solute}}$.
Since $P^{\circ} - P \propto X_{\text{solute}}$,we have $10 \propto 0.2$.
For a decrease of $20 \ mm \ Hg$,the new mole fraction of the solute $(X'_{\text{solute}})$ will be: $20 \propto X'_{\text{solute}}$.
Thus,$X'_{\text{solute}} = \frac{20}{10} \times 0.2 = 0.4$.
The mole fraction of the solvent is $X_{\text{solvent}} = 1 - X'_{\text{solute}} = 1 - 0.4 = 0.6$.

(A) According to Raoult's law for a binary solution of two volatile liquids $1$ and $2$:
$P_T = P_1^0 x_1 + P_2^0 x_2 = P_1^0 x_1 + P_2^0 (1 - x_1) = P_2^0 + x_1 (P_1^0 - P_2^0)$
Also,from Dalton's law,$P_1 = P_T y_1$,where $P_1 = P_1^0 x_1$.
So,$P_T y_1 = P_1^0 x_1 \implies P_T = \frac{P_1^0 x_1}{y_1}$.
Equating the two expressions for $P_T$:
$P_2^0 + x_1 (P_1^0 - P_2^0) = \frac{P_1^0 x_1}{y_1}$
Dividing both sides by $x_1 P_2^0$:
$\frac{P_2^0}{x_1} + \frac{P_1^0 - P_2^0}{P_2^0} = \frac{P_1^0}{P_2^0 y_1}$
Rearranging to the form $Y = mX + c$ where $Y = \frac{1}{x_1}$ and $X = \frac{1}{y_1}$:
$\frac{1}{x_1} = \left( \frac{P_1^0}{P_2^0} \right) \frac{1}{y_1} + \left( \frac{P_2^0 - P_1^0}{P_2^0} \right)$
Comparing with $Y = mX + c$,the slope is $\frac{P_1^0}{P_2^0}$ and the intercept is $\frac{P_2^0 - P_1^0}{P_2^0}$.

(D) According to Raoult's Law,the total vapour pressure of the solution is given by $P_{S} = P_{A}^{o} \cdot X_{A} + P_{B}^{o} \cdot X_{B}$.
Given: $n_{A} = 1 \ mole$,$n_{B} = 3 \ moles$,$P_{A}^{o} = 200 \ mm \ Hg$,$P_{S} = 500 \ mm \ Hg$.
Mole fractions are $X_{A} = \frac{1}{1+3} = 0.25$ and $X_{B} = \frac{3}{1+3} = 0.75$.
Substituting the values: $500 = (200 \times 0.25) + (P_{B}^{o} \times 0.75)$.
$500 = 50 + 0.75 \cdot P_{B}^{o}$.
$450 = 0.75 \cdot P_{B}^{o} \Rightarrow P_{B}^{o} = \frac{450}{0.75} = 600 \ mm \ Hg$.
Since $P_{A}^{o} (200 \ mm \ Hg) < P_{B}^{o} (600 \ mm \ Hg)$,component $A$ is the least volatile.

(D) Vapour pressure is an intensive property and is independent of the surface area of the liquid. It depends only on the nature of the liquid and the temperature. Therefore,the Assertion is false.
However,the rate of vaporisation is indeed directly proportional to the surface area of the liquid,as more molecules are exposed to the surface to escape into the vapour phase. Therefore,the Reason is true.

(C) The relative lowering of vapour pressure is given by the formula: $\frac{\Delta P}{P^o} = \frac{n_2}{n_1 + n_2} \approx \frac{n_2}{n_1} = m \times \frac{M_A}{1000}$.
Here,$\Delta P = 0.6 \ mm \ Hg$,$m = \frac{1}{18} \ mol \ kg^{-1}$,and $M_A$ (molar mass of water) $= 18 \ g \ mol^{-1}$.
Using the relation $\Delta P = P^o \times \frac{m \times M_A}{1000}$,we find the vapour pressure of pure solvent $P^o = \frac{\Delta P \times 1000}{m \times M_A} = \frac{0.6 \times 1000}{(1/18) \times 18} = 600 \ mm \ Hg$.
Since the solution is dilute,the vapour pressure of the solution $P_s = P^o - \Delta P = 600 - 0.6 = 599.4 \ mm \ Hg$.
However,in many textbook contexts,the approximation $\Delta P \approx P_s \times \frac{m \times M_A}{1000}$ is used,leading to $P_s = \frac{0.6 \times 1000}{18 \times (1/18)} = 600 \ mm \ Hg$.

(C) According to Raoult's law,the mole fraction of component $A$ in the vapour phase $(Y_A)$ is given by:
$Y_A = \frac{P_A}{P_{total}} = \frac{X_A P_A^{\circ}}{X_A P_A^{\circ} + X_B P_B^{\circ}}$
Given:
$P_A^{\circ} : P_B^{\circ} = 1 : 2 \implies P_B^{\circ} = 2 P_A^{\circ}$
$X_A : X_B = 1 : 2 \implies X_B = 2 X_A$
Substituting these values:
$Y_A = \frac{X_A P_A^{\circ}}{X_A P_A^{\circ} + (2 X_A)(2 P_A^{\circ})}$
$Y_A = \frac{X_A P_A^{\circ}}{X_A P_A^{\circ} + 4 X_A P_A^{\circ}}$
$Y_A = \frac{X_A P_A^{\circ}}{5 X_A P_A^{\circ}} = \frac{1}{5} = 0.2$

(C) According to Raoult's law for a solution containing a non-volatile solute,the relative lowering of vapour pressure is equal to the mole fraction of the solute.
Formula: $\frac{P^o - P_s}{P^o} = X_A$
Where:
$P^o = 0.70 \ atm$ (Vapour pressure of pure solvent)
$P_s = 0.35 \ atm$ (Vapour pressure of solution)
$X_A$ = Mole fraction of solute $A$
Substituting the values:
$X_A = \frac{0.70 - 0.35}{0.70}$
$X_A = \frac{0.35}{0.70}$
$X_A = 0.50$

(C) Given: $P_{benzene}^{\circ} = 160 \ torr$,$P_{toluene}^{\circ} = 60 \ torr$.
Since the solution is equimolar,the mole fractions are $x_{benzene} = 0.5$ and $x_{toluene} = 0.5$.
The partial pressure of benzene is $P_{benzene} = P_{benzene}^{\circ} \times x_{benzene} = 160 \times 0.5 = 80 \ torr$.
The partial pressure of toluene is $P_{toluene} = P_{toluene}^{\circ} \times x_{toluene} = 60 \times 0.5 = 30 \ torr$.
The total pressure is $P_{total} = P_{benzene} + P_{toluene} = 80 + 30 = 110 \ torr$.
The mole fraction of toluene in the vapour phase $(Y_{toluene})$ is given by $Y_{toluene} = \frac{P_{toluene}}{P_{total}} = \frac{30}{110} \approx 0.27$.

(D) According to Raoult's Law for a solution of two volatile liquids,$P_{total} = P_A^o x_A + P_B^o x_B$.
Given: $x_B = 0.4$,so $x_A = 1 - 0.4 = 0.6$.
Given: $P_A^o = 400 \ mm \ Hg$ and $P_{total} = 600 \ mm \ Hg$.
Substituting the values: $600 = (400 \times 0.6) + (P_B^o \times 0.4)$.
$600 = 240 + 0.4 P_B^o$.
$0.4 P_B^o = 600 - 240 = 360$.
$P_B^o = \frac{360}{0.4} = 900 \ mm \ Hg$.
Therefore,the correct option is $D$.

(D) According to Raoult's Law for a binary solution of two volatile liquids $A$ and $B$:
$P_{total} = P_A + P_B = P_A^0 \chi_A + P_B^0 \chi_B$
Given:
$P_B^0 = 900 \ mm \ Hg$
$\chi_B = 0.4$
$P_{total} = 600 \ mm \ Hg$
Since $\chi_A + \chi_B = 1$,we have $\chi_A = 1 - 0.4 = 0.6$.
Substituting the values into the equation:
$600 = P_A^0(0.6) + (900)(0.4)$
$600 = P_A^0(0.6) + 360$
$P_A^0(0.6) = 600 - 360 = 240$
$P_A^0 = \frac{240}{0.6} = 400 \ mm \ Hg$
Thus,the vapour pressure of pure liquid $A$ is $400 \ mm \ Hg$.

(B) According to Raoult's Law,the total vapour pressure of a solution is given by $P_{total} = P_A^0 x_A + P_B^0 x_B$.
Given: $P_{total} = 500 \ mmHg$,$P_A^0 = 400 \ mmHg$,$P_B^0 = 575 \ mmHg$.
Since $x_A + x_B = 1$,we can write $x_A = 1 - x_B$.
Substituting these values into the equation: $500 = 400(1 - x_B) + 575x_B$.
$500 = 400 - 400x_B + 575x_B$.
$500 - 400 = 175x_B$.
$100 = 175x_B$.
$x_B = \frac{100}{175} = \frac{4}{7} \approx 0.57$.

(C) According to Raoult's Law for non-volatile solutes: $\frac{P^0 - P}{P^0} = \frac{n_2}{n_1 + n_2} \approx \frac{n_2}{n_1} = \frac{W_2 \times M_1}{M_2 \times W_1}$
Given: $W_2 = 2 \ g$,$W_1 = 50 \ g$,$M_2 = 64 \ g \ mol^{-1}$,$M_1 = 78 \ g \ mol^{-1}$,$P^0 = 640 \ mm \ Hg$
$\frac{640 - P}{640} = \frac{2 \times 78}{64 \times 50}$
$\frac{640 - P}{640} = \frac{156}{3200} = 0.04875$
$640 - P = 640 \times 0.04875 = 31.2$
$P = 640 - 31.2 = 608.8 \ mm \ Hg$
The closest value is $608.64 \ mm \ Hg$.

(C) The mole fraction of $A$ is $x_A = \frac{n_A}{n_A + n_B} = \frac{2}{2+3} = 0.4$.
The mole fraction of $B$ is $x_B = \frac{n_B}{n_A + n_B} = \frac{3}{2+3} = 0.6$.
According to Raoult's law,the total vapour pressure of the solution is $P_{\text{total}} = P_{A}^{\circ} x_A + P_{B}^{\circ} x_B$.
Substituting the values: $P_{\text{total}} = (420 \times 0.4) + (610 \times 0.6)$.
$P_{\text{total}} = 168 + 366 = 534 \ mm \ Hg$.

(A) The relation between the vapour pressure of the solution $(P_1)$,the vapour pressure of the pure solvent $(P_1^*)$,and the mole fraction of the solvent $(x_1)$ in the solution is defined by Raoult's Law for an ideal solution.
Raoult's Law states that for a solution of volatile liquids,the partial vapour pressure of each component in the solution is directly proportional to its mole fraction present in the solution.
The mathematical expression is:
$P_1 = P_1^* \cdot x_1$
Where:
-$P_1$ is the partial vapour pressure of the solvent in the solution.
-$P_1^*$ is the vapour pressure of the pure solvent.
-$x_1$ is the mole fraction of the solvent in the solution.
Therefore,the correct option is $(A)$.

(D) Given: $n_2 = 1 \ mol$ (solute),$w_1 = 36 \ g$ (solvent),$P_1^0 = 32 \ mm \ Hg$.
Moles of water $(n_1)$ = $\frac{36 \ g}{18 \ g/mol} = 2 \ mol$.
According to Raoult's law for non-volatile solutes: $\frac{P_1^0 - P_1}{P_1^0} = \frac{n_2}{n_1 + n_2}$.
Substituting the values: $\frac{32 - P_1}{32} = \frac{1}{2 + 1} = \frac{1}{3}$.
$3(32 - P_1) = 32$.
$96 - 3P_1 = 32$.
$3P_1 = 96 - 32 = 64$.
$P_1 = \frac{64}{3} = 21.33 \ mm \ Hg$.

(A) Given: Vapour pressure of pure solvent,$P_1^0 = 240 \ mm Hg$.
Vapour pressure of solution,$P_1 = 216 \ mm Hg$.
According to Raoult's Law,the relative lowering of vapour pressure is equal to the mole fraction of the solute $(x_2)$:
$\frac{P_1^0 - P_1}{P_1^0} = x_2$
$\frac{240 - 216}{240} = x_2$
$x_2 = \frac{24}{240} = 0.1$
The mole fraction of the solvent $(x_1)$ is given by $x_1 = 1 - x_2$.
$x_1 = 1 - 0.1 = 0.9$.

(B) According to Raoult's law for non-volatile solutes: $\frac{P^{\circ} - P_s}{P^{\circ}} = X_{\text{solute}}$
Here,$P^{\circ} = 32 \ mm \ Hg$,$n_{\text{solute}} = 0.1 \ mol$.
The number of moles of solvent (water) is $n_{\text{solvent}} = \frac{16.2 \ g}{18 \ g/mol} = 0.9 \ mol$.
The mole fraction of the solute is $X_{\text{solute}} = \frac{n_{\text{solute}}}{n_{\text{solute}} + n_{\text{solvent}}} = \frac{0.1}{0.1 + 0.9} = \frac{0.1}{1.0} = 0.1$.
Substituting the values: $\frac{32 - P_s}{32} = 0.1$.
$32 - P_s = 32 \times 0.1 = 3.2$.
$P_s = 32 - 3.2 = 28.8 \ mm \ Hg$.

(D) The relative lowering of vapour pressure is given by the formula: $\frac{P_1^0 - P_s}{P_1^0} = X_{\text{solute}} = \frac{n_{\text{solute}}}{n_{\text{solute}} + n_{\text{solvent}}}$
Mass of water $= 1.8 \times 10^{-2} \ kg = 18 \ g$.
Moles of water $(n_{\text{solvent}})$ $= \frac{18 \ g}{18 \ g/mol} = 1 \ mol$.
Given $n_{\text{solute}} = 0.1 \ mol$ and $P_1^0 = 24 \ mm \ Hg$.
Substituting the values: $\frac{24 - P_s}{24} = \frac{0.1}{0.1 + 1} = \frac{0.1}{1.1} = \frac{1}{11}$.
$24 - P_s = \frac{24}{11} \approx 2.18 \ mm \ Hg$.
$P_s = 24 - 2.18 = 21.82 \ mm \ Hg \approx 21.84 \ mm \ Hg$.

(B) According to Raoult's law for non-volatile solutes: $\frac{P_1^0 - P_s}{P_1^0} = X_{\text{solute}} = \frac{n_{\text{solute}}}{n_{\text{solute}} + n_{\text{water}}}$.
Number of moles of water $(n_{\text{water}})$ = $\frac{36 \ g}{18 \ g/mol} = 2 \ mol$.
Given $n_{\text{solute}} = 1 \ mol$ and $P_1^0 = 400 \ mm \ Hg$.
Substituting the values: $\frac{400 - P_s}{400} = \frac{1}{1 + 2} = \frac{1}{3}$.
$400 - P_s = \frac{400}{3} = 133.33 \ mm \ Hg$.
$P_s = 400 - 133.33 = 266.67 \ mm \ Hg \approx 267 \ mm \ Hg$.

(C) The vapour pressure of a solution is determined by the mole fraction of the solvent. According to Raoult's Law,$P_{sol} = X_{solvent} \times P^0_{solvent}$.
Adding water to an aqueous solution increases the amount of solvent relative to the solute,thereby increasing the mole fraction of the solvent $(X_{solvent})$.
Since $P^0_{solvent}$ (vapour pressure of pure water) is constant at a given temperature,an increase in $X_{solvent}$ leads to an increase in the vapour pressure of the solution.
Adding more solute (as in options $A$,$B$,and $D$) would decrease the mole fraction of the solvent,thereby decreasing the vapour pressure.