Lowering of vapour pressure Questions in English

(B) According to Raoult's law for a solution containing a non-volatile solute,the relative lowering of vapor pressure is given by the equation: $\frac{P_1^0 - P_1}{P_1^0} = x_2$.
Here,$P_1^0$ is the vapor pressure of the pure solvent,$P_1$ is the vapor pressure of the solution,and $x_2$ is the mole fraction of the solute.
Thus,the relative lowering of vapor pressure is equal to the mole fraction of the solute.

(A) According to Raoult's law for non-volatile solutes,the relative lowering of vapor pressure is equal to the mole fraction of the solute.
$\frac{P^o - P_s}{P^o} = X_{solute}$
Given that the vapor pressure decreases by $8\%$,the relative lowering of vapor pressure is $\frac{P^o - P_s}{P^o} = \frac{8}{100} = 0.08$.
Therefore,the mole fraction of the solute $X_{solute} = 0.08$.

(B) At $373 \, K$,the vapor pressure of pure water $(P^0)$ is $1 \, atm$.
Given,the vapor pressure of the solution $(P_s)$ is $0.925 \, atm$.
According to Raoult's law,the relative lowering of vapor pressure is equal to the mole fraction of the solute $(x_2)$:
$\frac{P^0 - P_s}{P^0} = x_2$
Substituting the values:
$x_2 = \frac{1 - 0.925}{1} = 0.075$
Therefore,the mole fraction of the solute is $0.075$.

(A) The vapor pressure of a solution decreases as the concentration of solute particles (van't Hoff factor $i \times$ molarity $M$) increases.
Lower concentration of solute particles results in higher vapor pressure.
Let us calculate the concentration of particles for each:
$A$: $0.1 \ M$ Urea $(i=1)$ $\rightarrow 0.1 \times 1 = 0.1 \ M$ particles.
$B$: $0.2 \ M$ Sucrose $(i=1)$ $\rightarrow 0.2 \times 1 = 0.2 \ M$ particles.
$C$: $0.1 \ M \ CaCl_2$ $(i=3)$ $\rightarrow 0.1 \times 3 = 0.3 \ M$ particles.
$D$: $0.2 \ M \ KCl$ $(i=2)$ $\rightarrow 0.2 \times 2 = 0.4 \ M$ particles.
Since $0.1 \ M$ has the lowest concentration of solute particles,it will have the highest vapor pressure.

(C) According to Raoult's law,the relative lowering of vapor pressure is given by $\frac{P^o - P}{P^o} = \chi_{solute} = \frac{n_{solute}}{n_{solute} + n_{solvent}}$.
Since $n_{solute} \ll n_{solvent}$,the expression simplifies to $\frac{P^o - P}{P^o} \approx \frac{n_{solute}}{n_{solvent}} = \frac{x/M_{solute}}{y/M_{solvent}} = \frac{x \cdot M_{solvent}}{y \cdot M_{solute}}$.
For liquid $A$: $(\frac{P^o - P}{P^o})_A = \frac{x \cdot M_A}{y \cdot M_{solute}}$.
For liquid $B$: $(\frac{P^o - P}{P^o})_B = \frac{x \cdot M_B}{y \cdot M_{solute}}$.
Given that $(\frac{P^o - P}{P^o})_A = 2 \times (\frac{P^o - P}{P^o})_B$,we have $\frac{x \cdot M_A}{y \cdot M_{solute}} = 2 \times \frac{x \cdot M_B}{y \cdot M_{solute}}$.
Simplifying this,we get $M_A = 2 \ M_B$.

(D) According to Raoult's law for non-volatile solutes,the relative lowering of vapor pressure is equal to the mole fraction of the solute: $\frac{P^o - P_s}{P^o} = X_{solute}$.
Given,the decrease in vapor pressure $\Delta P = P^o - P_s = 11.5 \ torr$.
The mole fraction of the solute $X_{solute} = 0.2$.
Substituting these values into the equation: $\frac{11.5}{P^o} = 0.2$.
Therefore,$P^o = \frac{11.5}{0.2} = 57.5 \ torr$.
Thus,the vapor pressure of the pure solvent is $57.5 \ torr$.

(C) According to Raoult's Law,when a non-volatile solid solute is added to a volatile solvent,the solute particles occupy some surface area on the surface of the liquid.
This reduces the number of solvent molecules available at the surface to escape into the vapour phase.
Consequently,the rate of evaporation decreases,which leads to a decrease in the vapour pressure of the solution compared to the pure solvent.

(B) According to Raoult's Law for non-volatile solutes,the relative lowering of vapor pressure is given by: $\frac{P^o - P_s}{P^o} = X_{solute} = \frac{n}{n + N}$.
Here,$n = 1 \ mol$ (solute) and $N = 4 \ mol$ (solvent).
The mole fraction of the solute is $X_{solute} = \frac{1}{1 + 4} = \frac{1}{5} = 0.2$.
Given the vapor pressure of the solution $P_s = 24.0 \ kPa$.
Using the formula $\frac{P^o - 24.0}{P^o} = 0.2$,we get:
$P^o - 24.0 = 0.2 P^o$.
$0.8 P^o = 24.0$.
$P^o = \frac{24.0}{0.8} = 30 \ kPa$.

(A) According to Raoult's Law for a non-volatile solute,the relative lowering of vapor pressure is given by:
$\frac{P^o - P_s}{P^o} = \frac{w_2 \times M_1}{M_2 \times w_1}$
Given:
$P^o = 3000 \ N \ m^{-2}$
$P_s = 2985 \ N \ m^{-2}$
$w_2 = 5 \ g$ (mass of solute)
$w_1 = 100 \ g$ (mass of solvent,water)
$M_1 = 18 \ g/mol$ (molar mass of water)
Substituting the values:
$\frac{3000 - 2985}{3000} = \frac{5 \times 18}{M_2 \times 100}$
$\frac{15}{3000} = \frac{90}{M_2 \times 100}$
$0.005 = \frac{0.9}{M_2}$
$M_2 = \frac{0.9}{0.005} = 180 \ g/mol$
Therefore,the molar mass of the solute is $180 \ g/mol$.

(C) Given: Vapor pressure of pure water $(P^0_A) = 3000 \ N \ m^{-2}$.
Decrease in vapor pressure $(\Delta P) = 300 \ N \ m^{-2}$.
Vapor pressure of solution $(P_s) = P^0_A - \Delta P = 3000 - 300 = 2700 \ N \ m^{-2}$.
According to Raoult's law: $\frac{P^0_A - P_s}{P^0_A} = \frac{n_B}{n_A + n_B} \approx \frac{n_B}{n_A}$ (for dilute solutions).
$\frac{300}{3000} = \frac{n_B}{n_A} = 0.1$.
Here,$n_B$ is moles of solute and $n_A$ is moles of solvent (water).
$n_A = \frac{\text{mass of water}}{18} = \frac{1000 \ g}{18 \ g \ mol^{-1}} = 55.55 \ mol$.
Since $\frac{n_B}{n_A} = 0.1$,then $n_B = 0.1 \times 55.55 = 5.555 \ mol$.
Molality $(m) = \frac{n_B \times 1000}{\text{mass of solvent in grams}} = \frac{5.555 \times 1000}{1000} = 5.555 \ m \approx 5.55 \ m$.

(C) The vapor pressure of a solution decreases as the number of solute particles (van't Hoff factor,$i$) increases.
$1.$ Sugar is a non-electrolyte,so $i = 1$.
$2.$ $KCl$ dissociates as $K^+ + Cl^-$,so $i = 2$.
$3.$ $Cu(NO_3)_2$ dissociates as $Cu^{2+} + 2NO_3^-$,so $i = 3$.
$4.$ $AgNO_3$ dissociates as $Ag^+ + NO_3^-$,so $i = 2$.
Since $Cu(NO_3)_2$ produces the highest number of particles $(i = 3)$,it will cause the greatest lowering in vapor pressure.
Therefore,the solution with the lowest vapor pressure is $0.1 \, M \, Cu(NO_3)_2$.

(B) Given: Vapor pressure of pure water $(P^{\circ})$ = $19.8\,mm\,Hg$.
Mass of water $(w_A)$ = $178.2\,g$.
Molar mass of water $(M_A)$ = $18\,g/mol$.
Moles of water $(n_A)$ = $\frac{178.2}{18} = 9.9\,mol$.
Moles of glucose $(n_B)$ = $0.1\,mol$.
According to Raoult's Law,the relative lowering of vapor pressure is given by: $\frac{P^{\circ} - P_s}{P^{\circ}} = \frac{n_B}{n_A + n_B}$.
Substituting the values: $\frac{19.8 - P_s}{19.8} = \frac{0.1}{9.9 + 0.1} = \frac{0.1}{10} = 0.01$.
$19.8 - P_s = 19.8 \times 0.01 = 0.198$.
$P_s = 19.8 - 0.198 = 19.602\,mm\,Hg$.

(A) According to Raoult's Law,the relative lowering of vapor pressure is equal to the mole fraction of the solute $(x_2)$.
Formula: $\frac{P_1^o - P_s}{P_1^o} = x_2$
Given: $P_1^o = 0.80 \ atm$,$P_s = 0.60 \ atm$.
Substituting the values: $x_2 = \frac{0.80 - 0.60}{0.80} = \frac{0.20}{0.80} = 0.25$.
Therefore,the mole fraction of the solute is $0.25$.

(B) Assuming the previous context refers to Raoult's Law for a binary mixture of $A$ and $B$ where $P_A^0 = 450 \ mm \ Hg$ and $P_B^0 = 700 \ mm \ Hg$ (standard values for this specific problem type).
Given mole fraction of $A$ in liquid phase,$x_A = 0.8$.
Therefore,$x_B = 1 - 0.8 = 0.2$.
Partial pressure of $A$,$P_A = P_A^0 \times x_A = 450 \times 0.8 = 360 \ mm \ Hg$.
Partial pressure of $B$,$P_B = P_B^0 \times x_B = 700 \ \times 0.2 = 140 \ mm \ Hg$.
Total pressure,$P_{total} = P_A + P_B = 360 + 140 = 500 \ mm \ Hg$.
Mole fraction of $A$ in vapor phase,$y_A = \frac{P_A}{P_{total}} = \frac{360}{500} = 0.72$.
Since $0.72$ is closest to $0.7$,the correct option is $B$.

(B) According to Raoult's law,the partial pressure of $A$ is $P_A = x_2 P_A^o$.
In the vapor phase,the mole fraction $x_1$ is given by $x_1 = \frac{P_A}{P_{total}}$.
Substituting $P_A$,we get $x_1 = \frac{x_2 P_A^o}{P_{total}}$.
Rearranging for total pressure,$P_{total} = \frac{x_2}{x_1} P_A^o$.

(B) According to Raoult's Law,the total pressure $P$ of a binary solution is given by $P = P_A^0 X_A + P_B^0 X_B$.
Since $X_A + X_B = 1$,we have $X_B = 1 - X_A$.
Substituting this into the equation: $P = P_A^0 X_A + P_B^0 (1 - X_A) = (P_A^0 - P_B^0) X_A + P_B^0$.
Comparing this with the given equation $P = 0.172 X_A + 0.215$,we can see that the constant term represents the vapor pressure of pure $B$ $(P_B^0)$.
Therefore,$P_B^0 = 0.215 \ atm$.

(B) According to Raoult's law,the partial pressure of $A$ is $P_A = P_A^0 X_A$ and the total pressure is $P_{total} = P_A^0 X_A + P_B^0 X_B$.
In the vapor phase,the mole fraction $Y_A$ is given by $Y_A = \frac{P_A}{P_{total}} = \frac{P_A^0 X_A}{P_A^0 X_A + P_B^0 X_B}$.
Given that $P_A^0 < P_B^0$,the component with lower vapor pressure $(A)$ will be less volatile.
Therefore,the mole fraction of the less volatile component in the vapor phase is always less than its mole fraction in the liquid phase,i.e.,$Y_A < X_A$ or $X_A > Y_A$.

(C) According to Raoult's Law,the partial vapor pressure of a component in an ideal solution is given by the product of its mole fraction and its pure vapor pressure.
For component $A$,the partial vapor pressure $P_A$ is given by $P_A = X_A {P_A}^o$.
Since the sum of mole fractions in a binary solution is $X_A + X_B = 1$,we have $X_A = 1 - X_B$.
Substituting this into the equation,we get $P_A = (1 - X_B) {P_A}^o = {P_A}^o - X_B {P_A}^o$.

(D) According to Raoult's Law,the total vapor pressure $P$ of a binary mixture is given by $P = P_Y^0 + (P_X^0 - P_Y^0)X_x$.
Comparing this with the given equation $P = 160X_x + 50$:
$P_Y^0 = 50 \text{ torr}$ (vapor pressure of pure $Y$)
$P_X^0 - P_Y^0 = 160$
$P_X^0 - 50 = 160 \implies P_X^0 = 210 \text{ torr}$ (vapor pressure of pure $X$)
The ratio of vapor pressures of pure $X$ and $Y$ is $\frac{P_X^0}{P_Y^0} = \frac{210}{50} = \frac{21}{5} = 4.2 : 1$.

(C) Given: $p_{benzene}^o = 0.850 \ bar$,$p_{solution} = 0.845 \ bar$,$W_{benzene} = 39.0 \ g$,$M_{benzene} = 78 \ g/mol$,$w_{solid} = 0.5 \ g$.
Using Raoult's law for non-volatile solutes: $\frac{p^o - p}{p^o} = \frac{n_{solid}}{n_{solid} + n_{benzene}} \approx \frac{n_{solid}}{n_{benzene}}$ (for dilute solutions).
$n_{benzene} = \frac{39.0 \ g}{78 \ g/mol} = 0.5 \ mol$.
$\frac{0.850 - 0.845}{0.850} = \frac{0.5 / M_{solid}}{0.5}$.
$\frac{0.005}{0.850} = \frac{0.5}{M_{solid} \times 0.5} = \frac{1}{M_{solid}}$.
$M_{solid} = \frac{0.850}{0.005} = 170 \ g/mol$.

(C) For the initial solution: $n_A = 3, n_B = 2$. Total moles = $5$.
$P_1 = p_A^o \left( \frac{3}{5} \right) + p_B^o \left( \frac{2}{5} \right) = 600 \ torr$.
$3p_A^o + 2p_B^o = 3000$ (Equation $1$).
After adding $1.5 \ mol$ of $A$ and $0.5 \ mol$ of $C$: $n_A = 4.5, n_B = 2, n_C = 0.5$. Total moles = $7$.
Since $C$ is non-volatile,$p_C^o = 0$.
$P_2 = 600 + 30 = 630 \ torr$.
$P_2 = p_A^o \left( \frac{4.5}{7} \right) + p_B^o \left( \frac{2}{7} \right) = 630$.
$4.5p_A^o + 2p_B^o = 4410$ (Equation $2$).
Subtracting Equation $1$ from Equation $2$: $(4.5 - 3)p_A^o = 4410 - 3000 \implies 1.5p_A^o = 1410 \implies p_A^o = 940 \ torr$.
Substituting $p_A^o$ in Equation $1$: $3(940) + 2p_B^o = 3000 \implies 2820 + 2p_B^o = 3000 \implies 2p_B^o = 180 \implies p_B^o = 90 \ torr$.

(A) Moles of urea $= \frac{12 \ g}{60 \ g/mol} = 0.2 \ mol$.
Moles of sucrose $= \frac{68.4 \ g}{342 \ g/mol} = 0.2 \ mol$.
According to Raoult's law,the relative lowering of vapour pressure is equal to the mole fraction of the solute.
Since both solutions contain the same number of moles of solute in the same volume of solvent,their mole fractions are equal.
Therefore,the lowering of vapour pressure for both cases is equal.

(B) According to Raoult's law,the relative lowering of vapour pressure is equal to the mole fraction of the solute: $\frac{\Delta p}{p^o} = x_{solute}$.
Given $\Delta p = 10 \ mm \ Hg$ and $x_{solute} = 0.2$,we have $\frac{10}{p^o} = 0.2$,which gives $p^o = \frac{10}{0.2} = 50 \ mm \ Hg$.
For the second case,where $\Delta p = 20 \ mm \ Hg$,the new mole fraction of the solute $(x'_{solute})$ is $\frac{20}{p^o} = \frac{20}{50} = 0.4$.
Since the sum of mole fractions of solute and solvent is $1$,the mole fraction of the solvent is $x_{solvent} = 1 - x'_{solute} = 1 - 0.4 = 0.6$.

(C) According to Raoult's Law,the total vapour pressure of a solution containing two volatile components is given by $p_s = p_1 + p_2 = x_1 p_1^o + x_2 p_2^o$.
If the solute (component $2$) is more volatile than the solvent (component $1$),then $p_2^o > p_1^o$.
In this case,the total vapour pressure $p_s$ can be greater than the pure solvent vapour pressure $p_1^o$ depending on the mole fraction.
However,the Reason is incorrect because both components contribute to the total vapour pressure,and both form vapours,with the vapour phase being richer in the more volatile component.

(C) The pure solvent has a higher vapour pressure than the solution containing a non-volatile solute.
In the sealed container,the solvent molecules evaporate from the pure solvent beaker to establish equilibrium.
Since the solution has a lower vapour pressure,the solvent vapour condenses into the solution beaker to maintain the equilibrium vapour pressure.
Consequently,the volume of the pure solvent decreases while the volume of the solution increases.

(C) From the graph,at a constant vapour pressure,the temperature required for liquid $Z$ is the highest,followed by $Y$,and then $X$. This indicates that the boiling point order is $Z > Y > X$.
Liquids with higher boiling points have stronger intermolecular forces of attraction.
Therefore,the order of intermolecular interactions is $Z > Y > X$.
Evaluating the inferences:
$(A)$ $X$ has higher intermolecular interactions compared to $Y$: Incorrect,as $X$ has lower interactions than $Y$.
$(B)$ $X$ has lower intermolecular interactions compared to $Y$: Correct.
$(C)$ $Z$ has lower intermolecular interactions compared to $Y$: Incorrect,as $Z$ has higher interactions than $Y$.
Thus,only inference $(B)$ is correct.

$(i)$ Molar mass of $CH_{2}Cl_{2} = 12 \times 1 + 1 \times 2 + 35.5 \times 2 = 85\,g\,mol^{-1}$.
Molar mass of $CHCl_{3} = 12 \times 1 + 1 \times 1 + 35.5 \times 3 = 119.5\,g\,mol^{-1}$.
Moles of $CH_{2}Cl_{2} = \frac{40\,g}{85\,g\,mol^{-1}} = 0.47\,mol$.
Moles of $CHCl_{3} = \frac{25.5\,g}{119.5\,g\,mol^{-1}} = 0.213\,mol$.
Total number of moles $= 0.47 + 0.213 = 0.683\,mol$.
$x_{CH_{2}Cl_{2}} = \frac{0.47\,mol}{0.683\,mol} = 0.688$.
$x_{CHCl_{3}} = 1.00 - 0.688 = 0.312$.
$P_{\text{total}} = P_{CH_{2}Cl_{2}}^{0} \times x_{CH_{2}Cl_{2}} + P_{CHCl_{3}}^{0} \times x_{CHCl_{3}} = 415 \times 0.688 + 200 \times 0.312 = 285.52 + 62.4 = 347.92\,mm\,Hg$.
$(ii)$ Mole fraction in vapour phase $(y_i = P_i / P_{\text{total}})$.
$y_{CH_{2}Cl_{2}} = \frac{285.52}{347.92} = 0.82$.
$y_{CHCl_{3}} = \frac{62.4}{347.92} = 0.18$.

$(170 \ g \ mol^{-1}$ The given values are:
$p_{1}^{0} = 0.850 \ bar$; $p = 0.845 \ bar$; $M_{1} = 78 \ g \ mol^{-1}$; $w_{2} = 0.5 \ g$; $w_{1} = 39 \ g$
Using the relative lowering of vapour pressure formula:
$\frac{p_{1}^{0} - p}{p_{1}^{0}} = \frac{w_{2} \times M_{1}}{M_{2} \times w_{1}}$
Substituting the values:
$\frac{0.850 - 0.845}{0.850} = \frac{0.5 \times 78}{M_{2} \times 39}$
$\frac{0.005}{0.850} = \frac{39}{M_{2} \times 39}$
$\frac{1}{170} = \frac{1}{M_{2}}$
$M_{2} = 170 \ g \ mol^{-1}$

(A) Given:
$P_{A}^{0} = 450$ $mm$ $Hg$
$P_{B}^{0} = 700$ $mm$ $Hg$
$P_{total} = 600$ $mm$ $Hg$
According to Raoult's law:
$P_{total} = P_{A}^{0} x_{A} + P_{B}^{0} x_{B}$
Since $x_{B} = 1 - x_{A}$,we have:
$P_{total} = P_{A}^{0} x_{A} + P_{B}^{0} (1 - x_{A})$
$600 = 450 x_{A} + 700 (1 - x_{A})$
$600 = 450 x_{A} + 700 - 700 x_{A}$
$250 x_{A} = 100$
$x_{A} = \frac{100}{250} = 0.4$
$x_{B} = 1 - 0.4 = 0.6$
Composition of liquid mixture: $x_{A} = 0.4, x_{B} = 0.6$
Partial pressures:
$P_{A} = P_{A}^{0} x_{A} = 450 \times 0.4 = 180$ $mm$ $Hg$
$P_{B} = P_{B}^{0} x_{B} = 700 \times 0.6 = 420$ $mm$ $Hg$
Composition in vapour phase ($y_{A}$ and $y_{B}$):
$y_{A} = \frac{P_{A}}{P_{total}} = \frac{180}{600} = 0.3$
$y_{B} = \frac{P_{B}}{P_{total}} = \frac{420}{600} = 0.7$

(N/A) Given: Vapour pressure of pure water,$p_1^0 = 23.8 \, mm \, Hg$.
Weight of water,$w_1 = 850 \, g$.
Weight of urea,$w_2 = 50 \, g$.
Molar mass of water,$M_1 = 18 \, g \, mol^{-1}$.
Molar mass of urea,$M_2 = 60 \, g \, mol^{-1}$.
Number of moles of water,$n_1 = \frac{850}{18} = 47.22 \, mol$.
Number of moles of urea,$n_2 = \frac{50}{60} = 0.833 \, mol$.
Relative lowering of vapour pressure is given by $\frac{p_1^0 - p_1}{p_1^0} = \frac{n_2}{n_1 + n_2}$.
$\frac{p_1^0 - p_1}{p_1^0} = \frac{0.833}{47.22 + 0.833} = \frac{0.833}{48.053} = 0.0173$.
Now,$23.8 - p_1 = 0.0173 \times 23.8 = 0.4117$.
$p_1 = 23.8 - 0.4117 = 23.388 \, mm \, Hg \approx 23.4 \, mm \, Hg$.
Thus,the vapour pressure of the solution is $23.4 \, mm \, Hg$ and the relative lowering is $0.0173$.

(N/A) Given:
Vapour pressure of the solution at normal boiling point $(p_{1}) = 1.004 \ bar$
Vapour pressure of pure water at normal boiling point $(p_{1}^{\circ}) = 1.013 \ bar$
Mass of solute $(w_{2}) = 2 \ g$
Mass of solvent (water) $(w_{1}) = 98 \ g$
Molar mass of solvent (water) $(M_{1}) = 18 \ g \ mol^{-1}$
According to Raoult's law for relative lowering of vapour pressure:
$\frac{p_{1}^{\circ} - p_{1}}{p_{1}^{\circ}} = \frac{w_{2} \times M_{1}}{M_{2} \times w_{1}}$
Substituting the values:
$\frac{1.013 - 1.004}{1.013} = \frac{2 \times 18}{M_{2} \times 98}$
$\frac{0.009}{1.013} = \frac{36}{M_{2} \times 98}$
$M_{2} = \frac{36 \times 1.013}{0.009 \times 98}$
$M_{2} = \frac{36.468}{0.882} \approx 41.35 \ g \ mol^{-1}$
Thus,the molar mass of the solute is $41.35 \ g \ mol^{-1}$.

(N/A) $1 \, molal$ solution means $1 \, mol$ of the solute is present in $1000 \, g$ of the solvent (water).
Molar mass of water $= 18 \, g \, mol^{-1}$.
Number of moles of water $= \frac{1000 \, g}{18 \, g \, mol^{-1}} = 55.56 \, mol$.
Mole fraction of the solute $(x_2)$ $= \frac{n_2}{n_1 + n_2} = \frac{1}{55.56 + 1} = \frac{1}{56.56} \approx 0.0177$.
Given,vapour pressure of pure water,$p_1^o = 12.3 \, kPa$.
Using Raoult's Law for non-volatile solute: $\frac{p_1^o - p_1}{p_1^o} = x_2$.
$p_1 = p_1^o (1 - x_2) = 12.3 \times (1 - 0.0177) = 12.3 \times 0.9823$.
$p_1 \approx 12.08 \, kPa$.

(8 G) Let the vapour pressure of pure octane be $p_{1}^{o}$.
The final vapour pressure is $p_{1} = 0.8 p_{1}^{o}$.
Molar mass of solute,$M_{2} = 40 \, g \, mol^{-1}$.
Mass of octane,$w_{1} = 114 \, g$.
Molar mass of octane $(C_{8}H_{18})$,$M_{1} = (8 \times 12) + (18 \times 1) = 114 \, g \, mol^{-1}$.
Applying Raoult's Law for relative lowering of vapour pressure:
$\frac{p_{1}^{o} - p_{1}}{p_{1}^{o}} = \frac{w_{2} \times M_{1}}{M_{2} \times w_{1}}$
Substituting the values:
$\frac{p_{1}^{o} - 0.8 p_{1}^{o}}{p_{1}^{o}} = \frac{w_{2} \times 114}{40 \times 114}$
$\frac{0.2 p_{1}^{o}}{p_{1}^{o}} = \frac{w_{2}}{40}$
$0.2 = \frac{w_{2}}{40}$
$w_{2} = 0.2 \times 40 = 8 \, g$.
Hence,the required mass of the solute is $8 \, g$.

$(i)$ Let the molar mass of the solute be $M \,g \,mol^{-1}$.
Initial state:
$n_{1} = \frac{90 \,g}{18 \,g \,mol^{-1}} = 5 \,mol$ (moles of water)
$n_{2} = \frac{30 \,g}{M \,g \,mol^{-1}} = \frac{30}{M} \,mol$ (moles of solute)
$p_{1} = 2.8 \,kPa$
Using Raoult's Law: $\frac{p_{1}^{o} - p_{1}}{p_{1}^{o}} = \frac{n_{2}}{n_{1} + n_{2}}$
$1 - \frac{2.8}{p_{1}^{o}} = \frac{30/M}{5 + 30/M} = \frac{30}{5M + 30}$
$\frac{2.8}{p_{1}^{o}} = 1 - \frac{30}{5M + 30} = \frac{5M}{5M + 30}$
$\frac{p_{1}^{o}}{2.8} = \frac{5M + 30}{5M} \quad \dots (i)$
After adding $18 \,g$ of water:
$n_{1}' = \frac{90 + 18}{18} = 6 \,mol$
$p_{1}' = 2.9 \,kPa$
$\frac{p_{1}^{o} - 2.9}{p_{1}^{o}} = \frac{30/M}{6 + 30/M} = \frac{30}{6M + 30}$
$\frac{p_{1}^{o}}{2.9} = \frac{6M + 30}{6M} \quad \dots (ii)$
Dividing $(i)$ by $(ii)$:
$\frac{2.9}{2.8} = \frac{(5M + 30)/5M}{(6M + 30)/6M} = \frac{5M + 30}{5} \times \frac{6}{6M + 30} = \frac{6(5M + 30)}{5(6M + 30)}$
$2.9 \times 5(6M + 30) = 2.8 \times 6(5M + 30)$
$14.5(6M + 30) = 16.8(5M + 30)$
$87M + 435 = 84M + 504$
$3M = 69 \Rightarrow M = 23 \,g \,mol^{-1}$.
$(ii)$ Substituting $M = 23$ in $(i)$:
$\frac{p_{1}^{o}}{2.8} = \frac{5(23) + 30}{5(23)} = \frac{115 + 30}{115} = \frac{145}{115} \approx 1.2608$
$p_{1}^{o} = 2.8 \times 1.2608 = 3.53 \,kPa$.

(N/A) Vapour pressure of pure water,$P_{1}^{\circ} = 17.535 \, mm \, Hg$
Mass of glucose,$w_{2} = 25 \, g$
Mass of water,$w_{1} = 450 \, g$
Molar mass of glucose $(C_{6}H_{12}O_{6})$,$M_{2} = 6 \times 12 + 12 \times 1 + 6 \times 16 = 180 \, g \, mol^{-1}$
Molar mass of water,$M_{1} = 18 \, g \, mol^{-1}$
Number of moles of glucose,$n_{2} = \frac{25 \, g}{180 \, g \, mol^{-1}} = 0.139 \, mol$
Number of moles of water,$n_{1} = \frac{450 \, g}{18 \, g \, mol^{-1}} = 25 \, mol$
Using Raoult's law for relative lowering of vapour pressure:
$\frac{P_{1}^{\circ} - p_{1}}{P_{1}^{\circ}} = \frac{n_{2}}{n_{2} + n_{1}}$
$\frac{17.535 - p_{1}}{17.535} = \frac{0.139}{0.139 + 25} = \frac{0.139}{25.139} \approx 0.00553$
$17.535 - p_{1} = 0.00553 \times 17.535 \approx 0.097$
$p_{1} = 17.535 - 0.097 = 17.438 \, mm \, Hg \approx 17.44 \, mm \, Hg$
The vapour pressure of the solution is $17.44 \, mm \, Hg$.

(N/A) Number of moles of liquid $A$,$n_{A} = \frac{100}{140} \, mol = 0.714 \, mol$.
Number of moles of liquid $B$,$n_{B} = \frac{1000}{180} \, mol = 5.556 \, mol$.
Mole fraction of $A$,$x_{A} = \frac{n_{A}}{n_{A} + n_{B}} = \frac{0.714}{0.714 + 5.556} = 0.114$.
Mole fraction of $B$,$x_{B} = 1 - 0.114 = 0.886$.
Vapour pressure of pure liquid $B$,$P_{B}^{o} = 500 \, torr$.
Vapour pressure of liquid $B$ in the solution,$p_{B} = P_{B}^{o} x_{B} = 500 \times 0.886 = 443 \, torr$.
Total vapour pressure of the solution,$p_{total} = 475 \, torr$.
Vapour pressure of liquid $A$ in the solution,$p_{A} = p_{total} - p_{B} = 475 - 443 = 32 \, torr$.
Using Raoult's law,$p_{A} = P_{A}^{o} x_{A}$.
$P_{A}^{o} = \frac{p_{A}}{x_{A}} = \frac{32}{0.114} = 280.7 \, torr$.
Thus,the vapour pressure of pure liquid $A$ is $280.7 \, torr$ and its vapour pressure in the solution is $32 \, torr$.

(N/A) Consider a binary solution containing volatile components $1$ and $2$. Let their mole fractions be $X_{1}$ and $X_{2}$,and their partial vapour pressures be $P_{1}$ and $P_{2}$,respectively.
According to Raoult's law,for a solution of volatile liquids,the partial vapour pressure of each component is directly proportional to its mole fraction in the solution.
For component $1$: $P_{1} = P_{1}^{0} X_{1}$,where $P_{1}^{0}$ is the vapour pressure of pure component $1$.
For component $2$: $P_{2} = P_{2}^{0} X_{2}$,where $P_{2}^{0}$ is the vapour pressure of pure component $2$.
According to Dalton's law of partial pressures,the total pressure $P_{\text{total}}$ is the sum of the partial pressures:
$P_{\text{total}} = P_{1} + P_{2} = P_{1}^{0} X_{1} + P_{2}^{0} X_{2}$.
Since $X_{1} + X_{2} = 1$,we have $X_{1} = 1 - X_{2}$.
Substituting this:
$P_{\text{total}} = P_{1}^{0} (1 - X_{2}) + P_{2}^{0} X_{2} = P_{1}^{0} + (P_{2}^{0} - P_{1}^{0}) X_{2}$.
Conclusions:
$(i)$ $P_{\text{total}}$ is linearly related to the mole fraction of any one component.
$(ii)$ The graph of $P_{\text{total}}$ versus $X_{2}$ is a straight line.
$(iii)$ $P_{\text{total}}$ varies linearly with $X_{2}$ between $P_{1}^{0}$ and $P_{2}^{0}$.

(N/A) According to Raoult's law,the vapour pressure of a volatile component in a solution is given by $p_{i} = x_{i} p_{i}^{0}$.
In the case of a gas dissolved in a liquid,the gas is a volatile component. Its solubility is governed by Henry's law,which states that $P = K_{H} \cdot x$.
Comparing the two equations,we see that the partial pressure of the volatile component (or gas) is directly proportional to its mole fraction $(x)$ in the solution.
The only difference is the proportionality constant: $K_{H}$ in Henry's law and $p_{i}^{0}$ in Raoult's law.
Therefore,Raoult's law can be considered a special case of Henry's law where the proportionality constant $K_{H}$ becomes equal to the pure vapour pressure $p_{i}^{0}$.

(N/A) Liquids at a given temperature vaporize,and under equilibrium conditions,the pressure exerted by the vapours of the liquid over the liquid phase is called vapour pressure.
In a pure liquid,the entire surface is occupied by the molecules of the liquid. If a non-volatile solute is added to a solvent to form a solution,the vapour pressure of the solution is solely due to the solvent molecules.
This vapour pressure of the solution at a given temperature is found to be lower than the vapour pressure of the pure solvent at the same temperature. In the solution,the surface contains both solute and solvent molecules; thereby,the fraction of the surface covered by the solvent molecules is reduced. Consequently,the number of solvent molecules escaping from the surface is correspondingly reduced,thus,the vapour pressure is also reduced.

(N/A) According to Raoult's law,for a solution of two volatile components,the total pressure $(P_{total})$ is given by:
$P_{total} = P_1 + P_2$
Since $P_1 = x_1 P_1^0$ and $P_2 = x_2 P_2^0$,and $x_1 = 1 - x_2$:
$P_{total} = (1 - x_2) P_1^0 + x_2 P_2^0$
$P_{total} = P_1^0 - x_2 P_1^0 + x_2 P_2^0$
$P_{total} = P_1^0 + x_2(P_2^0 - P_1^0)$

(A) According to Raoult's Law,for an ideal solution,the partial vapour pressure of a component is directly proportional to its mole fraction in the solution.
Mathematically,$P_A = P_A^0 \times X_A$.
Here,$P_A$ is the partial vapour pressure of the solvent,$P_A^0$ is the vapour pressure of the pure solvent,and $X_A$ is the mole fraction of the solvent.
This equation follows the linear form $y = mx$,where $m = P_A^0$.
Therefore,the graph of $P_A$ versus $X_A$ is a straight line passing through the origin.

(A) $1$. Molar mass of benzene $(C_6H_6)$ = $78 \ g/mol$. Molar mass of toluene $(C_7H_8)$ = $92 \ g/mol$.
$2$. Moles of benzene $(n_B)$ = $7.8 \ g / 78 \ g/mol = 0.1 \ mol$.
$3$. Moles of toluene $(n_T)$ = $180 \ g / 92 \ g/mol \approx 1.9565 \ mol$.
$4$. Mole fraction of benzene $(x_B)$ = $0.1 / (0.1 + 1.9565) = 0.1 / 2.0565 \approx 0.0486$.
$5$. Mole fraction of toluene $(x_T)$ = $1 - 0.0486 = 0.9514$.
$6$. Total vapour pressure $(P_{total})$ = $P_B^0 \times x_B + P_T^0 \times x_T$.
$7$. $P_{total} = (0.9 \times 0.0486) + (0.85 \times 0.9514) = 0.04374 + 0.80869 = 0.85243 \ Bar$.

(A) $1$. Calculate the moles of each component:
$n_{\text{ethanol}} = \frac{60 \ g}{46 \ g/mol} \approx 1.304 \ mol$
$n_{\text{methanol}} = \frac{40 \ g}{32 \ g/mol} = 1.25 \ mol$
$2$. Calculate the mole fractions in the liquid phase:
$x_{\text{ethanol}} = \frac{1.304}{1.304 + 1.25} \approx 0.510$
$x_{\text{methanol}} = \frac{1.25}{1.304 + 1.25} \approx 0.490$
$3$. Calculate the partial pressures using Raoult's Law:
$P_{\text{ethanol}} = x_{\text{ethanol}} \times P^0_{\text{ethanol}} = 0.510 \times 44.5 \ mm \ Hg \approx 22.695 \ mm \ Hg$
$P_{\text{methanol}} = x_{\text{methanol}} \times P^0_{\text{methanol}} = 0.490 \times 88.7 \ mm \ Hg \approx 43.463 \ mm \ Hg$
$4$. Calculate the total pressure:
$P_{\text{total}} = 22.695 + 43.463 = 66.158 \ mm \ Hg$
$5$. Calculate the mole fraction of methanol in the vapour state $(y_{\text{methanol}})$:
$y_{\text{methanol}} = \frac{P_{\text{methanol}}}{P_{\text{total}}} = \frac{43.463}{66.158} \approx 0.657$
The closest value is $0.656$.

(B) Given: $P_{A}^{0} : P_{B}^{0} = 1 : 2$,so $P_{B}^{0} = 2P_{A}^{0}$.
Given: $X_{A} : X_{B} = 1 : 2$,so $X_{B} = 2X_{A}$.
Since $X_{A} + X_{B} = 1$,we have $X_{A} + 2X_{A} = 1$,which gives $3X_{A} = 1$,so $X_{A} = 1/3$ and $X_{B} = 2/3$.
According to Raoult's Law,the partial pressures are:
$P_{A} = X_{A} P_{A}^{0} = (1/3) P_{A}^{0}$
$P_{B} = X_{B} P_{B}^{0} = (2/3) (2P_{A}^{0}) = (4/3) P_{A}^{0}$
Total pressure $P_{T} = P_{A} + P_{B} = (1/3) P_{A}^{0} + (4/3) P_{A}^{0} = (5/3) P_{A}^{0}$.
The mole fraction of $A$ in the vapour state $(Y_{A})$ is given by $Y_{A} = P_{A} / P_{T}$.
$Y_{A} = ((1/3) P_{A}^{0}) / ((5/3) P_{A}^{0}) = 1/5 = 0.20$.

(0.450 BAR) According to Raoult's law,the total vapour pressure $P_{total} = P_A + P_B = x_A P_A^0 + x_B P_B^0$.
Given:
Moles of chlorobenzene $(n_A)$ = $0.1$
Moles of bromobenzene $(n_B)$ = $0.2$
Total moles = $0.1 + 0.2 = 0.3$
Mole fraction of chlorobenzene $(x_A)$ = $0.1 / 0.3 = 1/3$
Mole fraction of bromobenzene $(x_B)$ = $0.2 / 0.3 = 2/3$
$P_A^0 = 0.350 \ bar$
$P_B^0 = 0.500 \ bar$
$P_{total} = (1/3 \times 0.350) + (2/3 \times 0.500)$
$P_{total} = 0.1167 + 0.3333 = 0.450 \ bar$.