Lowering of vapour pressure Questions in English

(D) Given: Mole ratio $n_A : n_B = 1 : 4$. Therefore,mole fractions are $X_A = 1/5 = 0.2$ and $X_B = 4/5 = 0.8$.
Pure vapor pressures: $P_A^0 = 440 \ mm$,$P_B^0 = 120 \ mm$.
Total pressure $P_{total} = P_A^0 X_A + P_B^0 X_B = (440 \times 0.2) + (120 \times 0.8) = 88 + 96 = 184 \ mm$.
According to Dalton's Law,the mole fraction in the vapor phase $Y_A = P_A / P_{total}$.
$P_A = P_A^0 X_A = 440 \times 0.2 = 88 \ mm$.
$Y_A = 88 / 184 \approx 0.478$.

(B) According to Raoult's Law for relative lowering of vapour pressure:
$\frac{P^0 - P_s}{P_s} = \frac{n_2}{n_1} = \frac{w_2 \times M_1}{M_2 \times w_1}$
Given:
$P^0 = 640 \ mm \ Hg$,$P_s = 600 \ mm \ Hg$,$w_2 = 2.175 \ g$,$w_1 = 39.0 \ g$,$M_1 (\text{benzene, } C_6H_6) = 78 \ g/mol$
Substituting the values:
$\frac{640 - 600}{600} = \frac{2.175 \times 78}{M_2 \times 39.0}$
$\frac{40}{600} = \frac{2.175 \times 2}{M_2}$
$\frac{1}{15} = \frac{4.35}{M_2}$
$M_2 = 4.35 \times 15 = 65.25 \ g/mol$

(A) According to Raoult's Law for solutions containing non-volatile solutes,the relative lowering of vapor pressure is equal to the mole fraction of the solute. The mathematical expression is $\frac{P_1^o - P_1}{P_1^o} = x_2$,where $x_2$ is the mole fraction of the solute.

(D) According to Raoult's Law,the total vapor pressure of the solution is given by:
$P_{\text{total}} = P_A^0 X_A + P_B^0 X_B$
Given:
$P_A^0 = 80 \ mm$,$P_B^0 = 60 \ mm$
$n_A = 3 \ mol$,$n_B = 2 \ mol$
Total moles $= 3 + 2 = 5 \ mol$
Mole fraction of $A$ $(X_A)$ $= \frac{3}{5} = 0.6$
Mole fraction of $B$ $(X_B)$ $= \frac{2}{5} = 0.4$
$P_{\text{total}} = (80 \times 0.6) + (60 \times 0.4)$
$P_{\text{total}} = 48 + 24 = 72 \ mm$

(B) According to Raoult's Law for non-volatile solutes,the relative lowering of vapor pressure is equal to the mole fraction of the solute $(X_Y)$.
$P_{solvent}^0 = 0.80 \ atm$
$P_{solution} = 0.60 \ atm$
Relative lowering of vapor pressure = $\frac{P_{solvent}^0 - P_{solution}}{P_{solvent}^0} = X_Y$
$X_Y = \frac{0.80 - 0.60}{0.80} = \frac{0.20}{0.80} = \frac{1}{4} = 0.25$

(C) According to Raoult's Law,the total vapor pressure $P_{total}$ is given by:
$P_{total} = P_P^0 X_P + P_Q^0 X_Q$
Given:
$P_P^0 = 80 \ torr$,$P_Q^0 = 60 \ torr$
$n_P = 3 \ moles$,$n_Q = 2 \ moles$
Total moles $= 3 + 2 = 5 \ moles$
Mole fraction of $P$ $(X_P)$ $= \frac{3}{5} = 0.6$
Mole fraction of $Q$ $(X_Q)$ $= \frac{2}{5} = 0.4$
$P_{total} = (80 \times 0.6) + (60 \times 0.4)$
$P_{total} = 48 + 24 = 72 \ torr$

(B) Let the moles of pentane be $n_A = 1$ and moles of hexane be $n_B = 4$.
Total moles $= 1 + 4 = 5$.
Mole fraction of pentane in liquid phase,$X_A = \frac{1}{5} = 0.2$.
Mole fraction of hexane in liquid phase,$X_B = \frac{4}{5} = 0.8$.
Using Raoult's Law,partial pressure of pentane $P_A = P_A^0 \times X_A = 440 \times 0.2 = 88 \ mm \ Hg$.
Partial pressure of hexane $P_B = P_B^0 \times X_B = 120 \times 0.8 = 96 \ mm \ Hg$.
Total pressure $P_{total} = P_A + P_B = 88 + 96 = 184 \ mm \ Hg$.
Mole fraction of pentane in vapor phase $(Y_A)$ is given by $Y_A = \frac{P_A}{P_{total}} = \frac{88}{184} \approx 0.478$.

(D) The lowering of vapor pressure is directly proportional to the van't Hoff factor $(i)$ multiplied by the molar concentration $(C)$.
For a given concentration,the solution with the highest van't Hoff factor $(i)$ will have the greatest lowering of vapor pressure,and thus the lowest vapor pressure.
$i$ values are as follows:
$BaCl_2 \rightarrow Ba^{2+} + 2Cl^-$,$i = 3$
$\text{Urea} \rightarrow \text{non-electrolyte}$,$i = 1$
$Na_2SO_4 \rightarrow 2Na^+ + SO_4^{2-}$,$i = 3$
$Na_3PO_4 \rightarrow 3Na^+ + PO_4^{3-}$,$i = 4$
Since $Na_3PO_4$ has the highest $i$ value $(i=4)$,it causes the maximum lowering of vapor pressure,resulting in the lowest vapor pressure.

(C) The relative lowering of vapor pressure is a colligative property,which depends on the number of particles in the solution.
For equimolar solutions,the vapor pressure decreases as the van't Hoff factor $(i)$ increases.
$1$. $\text{Glucose}$ is a non-electrolyte,so $i = 1$.
$2$. $NaCl$ dissociates as $NaCl \rightarrow Na^+ + Cl^-$,so $i = 2$.
$3$. $Ba(NO_3)_2$ dissociates as $Ba(NO_3)_2 \rightarrow Ba^{2+} + 2NO_3^-$,so $i = 3$.
Since the number of particles is in the order $\text{Glucose} < NaCl < Ba(NO_3)_2$,the vapor pressure will be in the reverse order: $\text{Glucose} > NaCl > Ba(NO_3)_2$.

(C) Using the formula for relative lowering of vapour pressure: $\frac{P^0 - P_s}{P^0} = \frac{n_2}{n_1 + n_2} \approx \frac{n_2}{n_1} = \frac{W_2 / M_2}{W_1 / M_1}$
Given: $P^0 = 3000 \ N/m^2$,$P_s = 2985 \ N/m^2$,$W_2 = 5 \ g$,$W_1 = 100 \ g$,$M_1 = 18 \ g/mol$
$\frac{3000 - 2985}{3000} = \frac{5 / M_2}{100 / 18}$
$\frac{15}{3000} = \frac{5 \times 18}{100 \times M_2}$
$0.005 = \frac{90}{100 \times M_2}$
$M_2 = \frac{90}{100 \times 0.005} = \frac{90}{0.5} = 180 \ g/mol$

(A) Mass of $CCl_4 = \text{Volume} \times \text{Density} = 100\,mL \times 1.58\,g/mL = 158\,g$.
Molar mass of $CCl_4 (C=12, Cl=35.5) = 12 + 4 \times 35.5 = 154\,g/mol$.
Moles of solvent $(N_{solvent})$ $= 158 / 154 \approx 1.026\,mol$.
Moles of solute $(n_{solute})$ $= 0.5 / 65 \approx 0.00769\,mol$.
Using Raoult's Law: $(P^0 - P_s) / P_s = n_{solute} / N_{solvent}$.
$(143 - P_s) / P_s = (0.5 / 65) / (158 / 154)$.
$(143 - P_s) / P_s = (0.5 \times 154) / (65 \times 158) = 77 / 10270 \approx 0.0075$.
$143 - P_s = 0.0075 P_s \Rightarrow 143 = 1.0075 P_s$.
$P_s = 143 / 1.0075 \approx 141.93\,mm\,Hg$.

(A) According to Raoult's Law,the relative lowering of vapor pressure is equal to the mole fraction of the solute,but the ratio of the vapor pressure of the solution $(P_{solution})$ to the vapor pressure of the pure solvent $(P^0_{solvent})$ is equal to the mole fraction of the solvent $(X_{solvent})$.
$X_{solvent} = \frac{n_{solvent}}{n_{solvent} + n_{solute}}$
Given: $n_{solvent} = 2 \ mol$,$n_{solute} = 1 \ mol$
$\frac{P_{solution}}{P^0_{solvent}} = \frac{2}{2 + 1} = \frac{2}{3}$

(A) According to Raoult's law for a mixture of two volatile liquids,the total vapor pressure $P_s$ is given by:
$P_s = P_A^0 X_A + P_B^0 X_B$
Where $P_A^0$ is the vapor pressure of pure ethyl alcohol,$X_A = 0.6$ is its mole fraction,$P_B^0 = 200 \ mm$ is the vapor pressure of pure propyl alcohol,and $X_B = (1 - 0.6) = 0.4$ is its mole fraction.
Substituting the values:
$290 = P_A^0(0.6) + 200(0.4)$
$290 = 0.6 P_A^0 + 80$
$210 = 0.6 P_A^0$
$P_A^0 = \frac{210}{0.6} = 350 \ mm$.

(C) The relative lowering of vapour pressure is given by: $\frac{p^0 - p}{p^0} = \frac{n_2}{n_1 + n_2} \approx \frac{n_2}{n_1}$ (for dilute solutions).
Here,$p^0 = 25 \ mm$,$p = 24.5 \ mm$.
$\frac{25 - 24.5}{25} = \frac{0.5}{25} = 0.02$.
For a dilute aqueous solution,the relationship between relative lowering of vapour pressure and molality $(m)$ is: $\frac{p^0 - p}{p^0} = m \times \frac{M_{solvent}}{1000}$.
Given $M_{water} = 18 \ g/mol$,we have $0.02 = m \times \frac{18}{1000}$.
$m = \frac{0.02 \times 1000}{18} = \frac{20}{18} = 1.11 \ m$.

(B) Given that the vapour pressure of the solution is $2\%$ less than that of pure water,let $P^o = 100 \text{ bar}$,then $P_s = 98 \text{ bar}$.
According to Raoult's Law for a non-volatile solute,the relative lowering of vapour pressure is given by:
$\frac{P^o - P_s}{P_s} = \frac{n}{N} = \frac{w \times M}{m \times W}$
Where $w/m$ is the number of moles of solute and $W$ is the mass of solvent in grams.
Molality $(m') = \frac{w \times 1000}{m \times W}$.
Substituting this into the equation:
$\frac{100 - 98}{98} = m' \times \frac{M_{solvent}}{1000}$
$\frac{2}{98} = m' \times \frac{18}{1000}$
$m' = \frac{2 \times 1000}{98 \times 18} = \frac{2000}{1764} \approx 1.133 \text{ mol/kg}$.

(B) The vapor pressure of a liquid is inversely proportional to its boiling point.
As the boiling point increases,the intermolecular forces of attraction increase,leading to a decrease in vapor pressure.
Given boiling points:
$C_6H_6 = 80 \, ^\circ C$
$CH_3OH = 65 \, ^\circ C$
$C_6H_5NH_2 = 184 \, ^\circ C$
$C_6H_5NO_2 = 212 \, ^\circ C$
Since $CH_3OH$ has the lowest boiling point $(65 \, ^\circ C)$,it will have the highest vapor pressure at room temperature.

(C) Let the initial vapour pressure of pure solvent be $P_0 = 100 \ mm \ Hg$.
Given that the decrease in vapour pressure is $2.5 \%$,the lowering of vapour pressure is $\Delta P = 2.5 \ mm \ Hg$.
Thus,the vapour pressure of the solution is $P_S = P_0 - \Delta P = 100 - 2.5 = 97.5 \ mm \ Hg$.
Using Raoult's Law for relative lowering of vapour pressure: $\frac{P_0 - P_S}{P_S} = \frac{n}{N}$,where $n$ is the moles of solute and $N$ is the moles of solvent.
$n = \frac{W}{90}$ and $N = \frac{97.5}{18}$.
Substituting the values: $\frac{2.5}{97.5} = \frac{W / 90}{97.5 / 18}$.
$\frac{2.5}{97.5} = \frac{W \times 18}{90 \times 97.5}$.
$2.5 = \frac{W \times 18}{90} = \frac{W}{5}$.
$W = 2.5 \times 5 = 12.5 \ g$.

(A) According to $Raoult's$ law,for a solution containing a volatile solvent,the partial vapor pressure of the solvent $(p_1)$ is directly proportional to its mole fraction $(x_1)$ in the solution.
Mathematically,$p_1 \propto x_1$ or $p_1 = p_1^0 \cdot x_1$,where $p_1^0$ is the vapor pressure of the pure solvent.
Therefore,the vapor pressure of the solvent is proportional to the mole fraction of the solvent.

(C) The vapour pressure of pure water at $100 \ ^\circ C$ is $P^0 = 760 \ torr$.
The number of moles of glucose (solute) is $n = \frac{18 \ g}{180 \ g/mol} = 0.1 \ mol$.
The number of moles of water (solvent) is $N = \frac{178.2 \ g}{18 \ g/mol} = 9.9 \ mol$.
Using Raoult's law for relative lowering of vapour pressure: $\frac{P^0 - P_s}{P^0} = \frac{n}{n + N}$.
$\frac{760 - P_s}{760} = \frac{0.1}{0.1 + 9.9} = \frac{0.1}{10} = 0.01$.
$760 - P_s = 760 \times 0.01 = 7.6$.
$P_s = 760 - 7.6 = 752.4 \ torr$.

(C) According to Raoult's Law,the partial pressure of components are: $P_A = X_A \times P_A^o = 0.4 \times 100 = 40 \, mm \, Hg$.
$P_B = X_B \times P_B^o = (1 - 0.4) \times 200 = 0.6 \times 200 = 120 \, mm \, Hg$.
Total pressure $P_{total} = P_A + P_B = 40 + 120 = 160 \, mm \, Hg$.
The mole fraction of $A$ in the vapor phase $(Y_A)$ is given by: $Y_A = \frac{P_A}{P_{total}} = \frac{40}{160} = 0.25$.

(C) According to Raoult's law for non-volatile solutes:
$(1)$ The decrease in vapour pressure $(\Delta P = P_A^o - P_A)$ is equal to $P_A^o \times \chi_B$. This is not equal to the mole fraction of the solute $(\chi_B)$. So,statement $(1)$ is $F$.
$(2)$ The relative lowering of vapour pressure is $\frac{P_A^o - P_A}{P_A^o} = \chi_B$. Since $\chi_B = \frac{n_B}{n_A + n_B}$,for dilute solutions,it is proportional to the amount of solute $(n_B)$. So,statement $(2)$ is $T$.
$(3)$ The relative lowering of vapour pressure is equal to the mole fraction of the solute $(\chi_B)$. So,statement $(3)$ is $T$.
$(4)$ The vapour pressure of the solution is $P_A = P_A^o \times \chi_A$. It is not equal to the mole fraction of the solvent $(\chi_A)$ unless $P_A^o = 1$. So,statement $(4)$ is $F$.
Thus,the sequence is $FTTF$.

(A) Given: Density of $CCl_4$ $(d)$ $= 1.58 \ g/cm^3$,Volume of $CCl_4$ $(V)$ $= 100 \ mL$.
Mass of $CCl_4$ $(W_0)$ $= d \times V = 1.58 \times 100 = 158 \ g$.
Molar mass of $CCl_4$ $(M_0)$ $= 12 + 4 \times 35.5 = 154 \ g/mol$.
Moles of $CCl_4$ $(N)$ $= W_0 / M_0 = 158 / 154 \approx 1.026 \ mol$.
Mass of solute $(W)$ $= 0.5 \ g$,Molar mass of solute $(M)$ $= 65 \ g/mol$.
Moles of solute $(n)$ $= W / M = 0.5 / 65 \approx 0.00769 \ mol$.
Using Raoult's Law for dilute solutions: $(p^0 - p) / p^0 = n / N$.
$(143 - p) / 143 = 0.00769 / 1.026$.
$(143 - p) / 143 \approx 0.007495$.
$143 - p = 143 \times 0.007495 \approx 1.0718$.
$p = 143 - 1.0718 = 141.9282 \approx 141.93 \ mm \ Hg$.

(A) According to Raoult's law for a solution containing a non-volatile solid solute and a volatile liquid solvent,the vapor pressure of the solution $(P_{sol})$ is given by the equation: $P_{sol} = P_A^0 \times x_A$,where $P_A^0$ is the vapor pressure of the pure solvent and $x_A$ is the mole fraction of the solvent.
Therefore,the vapor pressure of the solution is directly proportional to the mole fraction of the solvent.

(D) According to Raoult's law for a solution containing a non-volatile solute,the relative lowering of vapour pressure is given by the expression: $\frac{P_1^o - P_1}{P_1^o} = x_2$.
Here,$\frac{P_1^o - P_1}{P_1^o}$ represents the relative lowering of vapour pressure,and $x_2$ represents the mole fraction of the solute.
Therefore,the relative lowering of vapour pressure is equal to the mole fraction of the solute.

(B) $1 \, molal$ solution means $1 \, mole$ of solute is dissolved in $1000 \, g$ of solvent (water).
Moles of solvent $(N)$ = $\frac{1000 \, g}{18 \, g/mol} = 55.56 \, mol$.
Moles of solute $(n)$ = $1 \, mol$.
According to Raoult's Law for a non-volatile solute:
$P_s = X_{solvent} \times P^o$
$P_s = \left( \frac{N}{n + N} \right) \times P^o$
Substituting the values:
$P_s = \left( \frac{55.56}{1 + 55.56} \right) \times 12.3 \, kPa$
$P_s = \left( \frac{55.56}{56.56} \right) \times 12.3 \, kPa$
$P_s = 0.9823 \times 12.3 \, kPa \approx 12.08 \, kPa$.

(B) Using the relative lowering of vapour pressure formula:
$\frac{P^o - P_s}{P_s} = \frac{w \times M}{m \times W}$
Where:
$P^o = 1020 \, \text{torr}$ (Vapour pressure of pure benzene)
$P_s = 990 \, \text{torr}$ (Vapour pressure of solution)
$w = 5 \, g$ (Mass of solute)
$W = 58.5 \, g$ (Mass of solvent,benzene)
$M = 78 \, g/mol$ (Molar mass of benzene)
$m$ = Molar mass of solute
Substituting the values:
$\frac{1020 - 990}{990} = \frac{5 \times 78}{m \times 58.5}$
$\frac{30}{990} = \frac{390}{m \times 58.5}$
$\frac{1}{33} = \frac{390}{m \times 58.5}$
$m = \frac{390 \times 33}{58.5} = 220 \, g/mol$

(C) The decrease in vapor pressure $(V.P.)$ is proportional to the van't Hoff factor $(i)$ multiplied by the molar concentration $(C)$.
For $0.1 \ M \ KCl$,$i = 2$,so $C \times i = 0.1 \times 2 = 0.2$.
For $0.1 \ M$ Urea,$i = 1$,so $C \times i = 0.1 \times 1 = 0.1$.
For $0.1 \ M \ BaCl_2$,$i = 3$,so $C \times i = 0.1 \times 3 = 0.3$.
For $0.1 \ M \ NaCl$,$i = 2$,so $C \times i = 0.1 \times 2 = 0.2$.
Since $0.1 \ M \ BaCl_2$ has the highest value of $C \times i$,it causes the maximum decrease in the vapor pressure of water.

(A) According to Raoult's Law for an ideal solution,$P_s = P_X^o X_X + P_Y^o X_Y$.
Initially,$n_X = 1$ and $n_Y = 3$,so $X_X = 1/4$ and $X_Y = 3/4$.
$550 = P_X^o (1/4) + P_Y^o (3/4) \Rightarrow P_X^o + 3P_Y^o = 2200$ (Equation $1$).
After adding $1 \ mol$ of $Y$,$n_X = 1$ and $n_Y = 4$,so $X_X = 1/5$ and $X_Y = 4/5$.
The new vapor pressure is $550 + 10 = 560 \ mm \ Hg$.
$560 = P_X^o (1/5) + P_Y^o (4/5) \Rightarrow P_X^o + 4P_Y^o = 2800$ (Equation $2$).
Subtracting Equation $1$ from Equation $2$:
$(P_X^o + 4P_Y^o) - (P_X^o + 3P_Y^o) = 2800 - 2200 \Rightarrow P_Y^o = 600 \ mm \ Hg$.
Substituting $P_Y^o = 600$ into Equation $1$:
$P_X^o + 3(600) = 2200$ $\Rightarrow P_X^o + 1800 = 2200$ $\Rightarrow P_X^o = 400 \ mm \ Hg$.

(D) Given:
$P^o = 200 \ mm \ Hg$
$P_s = 195 \ mm \ Hg$
$w = 2 \ g$ (mass of solute)
$W = 78 \ g$ (mass of solvent,benzene)
$M = 78 \ g/mol$ (molar mass of benzene,$C_6H_6$)
Using Raoult's law for non-volatile solute:
$\frac{P^o - P_s}{P^o} = \frac{w \times M}{m \times W}$
$\frac{200 - 195}{200} = \frac{2 \times 78}{m \times 78}$
$\frac{5}{200} = \frac{2}{m}$
$m = \frac{2 \times 200}{5} = 80 \ g/mol$
Therefore,the molecular weight of the solid is $80 \ g/mol$.

(D) $1$. Calculate the moles of each component:
$n_{CHCl_3} = \frac{25.5 \ g}{119.5 \ g/mol} \approx 0.213 \ mol$
$n_{CH_2Cl_2} = \frac{40 \ g}{85 \ g/mol} \approx 0.471 \ mol$
$2$. Calculate the mole fractions:
$n_{total} = 0.213 + 0.471 = 0.684 \ mol$
$x_{CHCl_3} = \frac{0.213}{0.684} \approx 0.311$
$x_{CH_2Cl_2} = \frac{0.471}{0.684} \approx 0.689$
$3$. Calculate the total vapour pressure using Raoult's Law:
$P_{total} = P^0_{CHCl_3} \cdot x_{CHCl_3} + P^0_{CH_2Cl_2} \cdot x_{CH_2Cl_2}$
$P_{total} = (200 \ mm \ Hg \times 0.311) + (41.5 \ mm \ Hg \times 0.689)$
$P_{total} = 62.2 + 28.5935 = 90.7935 \ mm \ Hg$
Since $90.7935 \ mm \ Hg$ is not among the options,the correct choice is $D$.

(D) The vapour pressure of a solution is inversely proportional to the concentration of the solute particles.
Adding water to an aqueous solution of $KI$ dilutes the solution,which decreases the concentration of solute particles.
According to Raoult's law,a decrease in solute concentration leads to an increase in the vapour pressure of the solvent.
In contrast,adding $NaCl$,$Na_2SO_4$,or more $KI$ increases the total number of solute particles,which further lowers the vapour pressure due to the colligative property of vapour pressure lowering.

(B) Moles of glucose $(n_{glucose}) = \frac{18 \ g}{180 \ g/mol} = 0.1 \ mol$.
Moles of water $(n_{water}) = \frac{178.2 \ g}{18 \ g/mol} = 9.9 \ mol$.
Mole fraction of water $(\chi_{water}) = \frac{n_{water}}{n_{water} + n_{glucose}} = \frac{9.9}{9.9 + 0.1} = \frac{9.9}{10} = 0.99$.
The vapour pressure of pure water at $100^\circ C$ is $760 \ Torr$.
According to Raoult's law,the vapour pressure of the solution $(P_{sol}) = \chi_{water} \times P^\circ_{water}$.
$P_{sol} = 0.99 \times 760 \ Torr = 752.4 \ Torr$.

(B) According to Raoult's law,the total vapour pressure of the solution is given by $P = P_A^\circ x_A + P_B^\circ x_B$.
Given:
$P = 290 \, mm$
$P_B^\circ = 200 \, mm$ (vapour pressure of propyl alcohol)
$x_A = 0.6$ (mole fraction of ethyl alcohol)
$x_B = 1 - x_A = 1 - 0.6 = 0.4$ (mole fraction of propyl alcohol)
Substituting the values in the equation:
$290 = P_A^\circ \times 0.6 + 200 \times 0.4$
$290 = P_A^\circ \times 0.6 + 80$
$290 - 80 = P_A^\circ \times 0.6$
$210 = P_A^\circ \times 0.6$
$P_A^\circ = \frac{210}{0.6} = 350 \, mm$.

(D) At $1\, atm$ atmospheric pressure,the boiling point of the mixture is $80\,^oC.$
At the boiling point,the total vapour pressure of the mixture,$P_{T} = 1\, atm = 760\, mm\,Hg.$
Using Raoult's law,$P_{T} = P_{A}^{\circ} X_{A} + P_{B}^{\circ} X_{B}.$
Given $P_{A}^{\circ} = 520\, mm\,Hg,$ $P_{B}^{\circ} = 1000\, mm\,Hg,$ and $X_{A} + X_{B} = 1,$
$760 = 520 X_{A} + 1000(1 - X_{A}).$
$760 = 520 X_{A} + 1000 - 1000 X_{A}.$
$480 X_{A} = 240.$
$X_{A} = \frac{240}{480} = 0.5.$
Therefore,the amount of $'A'$ in the mixture is $50\, mol\, percent.$

(A) The relative lowering of vapour pressure is given by Raoult's Law: $\frac{P^o - P_S}{P^o} = \frac{n_2}{n_1 + n_2} \approx \frac{n_2}{n_1}$ for dilute solutions.
$1.$ Calculate moles of glucose $(n_2)$: $n_2 = \frac{18 \, g}{180 \, g/mol} = 0.1 \, mol$.
$2.$ Calculate moles of water $(n_1)$: $n_1 = \frac{178.2 \, g}{18 \, g/mol} = 9.9 \, mol$.
$3.$ Using the formula $\frac{P^o - P_S}{P^o} = \frac{n_2}{n_1 + n_2}$:
$\frac{17.5 - P_S}{17.5} = \frac{0.1}{9.9 + 0.1} = \frac{0.1}{10} = 0.01$.
$4.$ Solve for $P_S$:
$17.5 - P_S = 17.5 \times 0.01 = 0.175$.
$P_S = 17.5 - 0.175 = 17.325 \, mm \, Hg$.

(D) The relative lowering of vapour pressure is given by the formula: $\frac{P^o - P_s}{P_s} = \frac{w_2 \times M_1}{w_1 \times M_2}$
Here,$P^o = 185\,torr$ (vapour pressure of pure solvent),
$P_s = 183\,torr$ (vapour pressure of solution),
$w_1 = 100\,g$ (mass of solvent,acetone),
$M_1 = 58\,g\,mol^{-1}$ (molar mass of acetone,$CH_3COCH_3$),
$w_2 = 1.2\,g$ (mass of solute),
$M_2 = ?$ (molar mass of solute).
Substituting the values:
$\frac{185 - 183}{183} = \frac{1.2 \times 58}{100 \times M_2}$
$\frac{2}{183} = \frac{69.6}{100 \times M_2}$
$M_2 = \frac{69.6 \times 183}{2 \times 100} = \frac{12736.8}{200} = 63.684 \approx 64\,g\,mol^{-1}$.

(A) According to Raoult's Law for non-volatile solutes:
$\frac{P^{\circ} - P_s}{P^{\circ}} = \frac{n_B}{n_A + n_B}$
Given:
Mass of glucose $(W_B) = 18 \ g$,Molar mass $(M_B) = 180 \ g/mol$
Moles of glucose $(n_B) = \frac{18}{180} = 0.1 \ mol$
Mass of water $(W_A) = 178.2 \ g$,Molar mass $(M_A) = 18 \ g/mol$
Moles of water $(n_A) = \frac{178.2}{18} = 9.9 \ mol$
Vapour pressure of pure water at $100 \ ^{\circ}C$ $(P^{\circ}) = 760 \ torr$
Substituting the values:
$\frac{760 - P_s}{760} = \frac{0.1}{9.9 + 0.1} = \frac{0.1}{10} = 0.01$
$760 - P_s = 760 \times 0.01 = 7.6$
$P_s = 760 - 7.6 = 752.4 \ torr$

(C) According to Raoult's Law,the total vapour pressure of the solution is given by $P_{T} = P^{\circ}_{A} X_{A} + P^{\circ}_{B} X_{B}$.
First,calculate the number of moles of $A$ and $B$:
$n_{A} = \frac{20\, g}{40\, g/mol} = 0.5\, mol$
$n_{B} = \frac{20\, g}{80\, g/mol} = 0.25\, mol$
Next,calculate the mole fractions:
$X_{A} = \frac{n_{A}}{n_{A} + n_{B}} = \frac{0.5}{0.5 + 0.25} = \frac{0.5}{0.75} = \frac{2}{3}$
$X_{B} = \frac{n_{B}}{n_{A} + n_{B}} = \frac{0.25}{0.75} = \frac{1}{3}$
Finally,calculate the total vapour pressure:
$P_{T} = (100\, torr \times \frac{2}{3}) + (40\, torr \times \frac{1}{3})$
$P_{T} = \frac{200}{3} + \frac{40}{3} = \frac{240}{3} = 80\, torr$.

(B) The relative lowering of vapour pressure is given by the formula: $\frac{P^{\circ} - P_s}{P_s} = \frac{n_B}{n_A}$.
Here,$P^{\circ} = 92.5 \ mm$ is the vapour pressure of pure water.
For a $1 \ molal$ solution,$n_B = 1 \ mol$ of solute is dissolved in $1000 \ g$ of water.
The number of moles of water $(n_A)$ is $\frac{1000 \ g}{18 \ g/mol} = 55.55 \ mol$.
Substituting the values: $\frac{92.5 - P_s}{P_s} = \frac{1}{55.55}$.
$92.5 - P_s = \frac{P_s}{55.55}$.
$92.5 = P_s (1 + 0.018) = 1.018 \ P_s$.
$P_s = \frac{92.5}{1.018} \approx 90.86 \ mm$.

(B) Let the total mass of the solution be $100 \ g$.
Mass of liquid $A = 28 \ g$,Mass of water $= 72 \ g$.
Moles of $A$ $(n_A)$ $= \frac{28}{140} = 0.2 \ mol$.
Moles of water $(n_w)$ $= \frac{72}{18} = 4 \ mol$.
Total moles $= 0.2 + 4 = 4.2 \ mol$.
Mole fraction of water $(x_w)$ $= \frac{4}{4.2} = \frac{20}{21}$.
Mole fraction of $A$ $(x_A)$ $= \frac{0.2}{4.2} = \frac{1}{21}$.
According to Raoult's Law for a solution of two volatile liquids: $P_{total} = P_w^\circ x_w + P_A^\circ x_A$.
$160 = 150 \times (\frac{20}{21}) + P_A^\circ \times (\frac{1}{21})$.
$160 = \frac{3000}{21} + \frac{P_A^\circ}{21}$.
$3360 = 3000 + P_A^\circ$.
$P_A^\circ = 360 \ mm$.

(A) According to Raoult's Law for non-volatile solutes: $\frac{P^{\theta} - P}{P^{\theta}} = \frac{n_2}{n_1 + n_2}$.
Given that the vapour pressure is reduced to $80 \%$ of the pure octane,$P = 0.8 P^{\theta}$,so $\frac{P^{\theta} - 0.8 P^{\theta}}{P^{\theta}} = 0.2$.
For octane $(C_8H_{18})$,molar mass $M_1 = (8 \times 12) + (18 \times 1) = 114 \ g/mol$.
Number of moles of octane $n_1 = \frac{114 \ g}{114 \ g/mol} = 1 \ mol$.
Number of moles of solute $n_2 = \frac{x}{40}$.
Substituting into the equation: $0.2 = \frac{n_2}{n_1 + n_2} = \frac{x/40}{1 + x/40}$.
$0.2(1 + x/40) = x/40 \Rightarrow 0.2 + 0.2(x/40) = x/40$.
$0.2 = 0.8(x/40) \Rightarrow 0.2 = x/50$.
$x = 0.2 \times 50 = 10 \ g$.

(A) $1$. Calculate the moles of acetone $(CH_3COCH_3)$: Molar mass = $58 \ g/mol$. Moles = $174 \ g / 58 \ g/mol = 3 \ mol$.
$2$. Calculate the moles of water $(H_2O)$: Molar mass = $18 \ g/mol$. Moles = $126 \ g / 18 \ g/mol = 7 \ mol$.
$3$. Calculate the mole fractions: Total moles = $3 + 7 = 10 \ mol$. $X_{acetone} = 3/10 = 0.3$. $X_{H_2O} = 7/10 = 0.7$.
$4$. Apply Raoult's Law: $P_{total} = P^0_{acetone} \times X_{acetone} + P^0_{H_2O} \times X_{H_2O}$.
$5$. $P_{total} = (360 \ torr \times 0.3) + (24 \ torr \times 0.7) = 108 + 16.8 = 124.8 \ torr$.

(A) The loss in weight of pure water is proportional to the vapour pressure of pure water $(P^0)$,and the loss in weight of the solution is proportional to the vapour pressure of the solution $(P_s)$.
Let the loss in weight of pure water be $w_1 = 1.5 + 2 = 3.5 \ g$.
Let the loss in weight of the solution be $w_2 = 2 \ g$.
According to the relative lowering of vapour pressure formula: $\frac{P^0 - P_s}{P^0} = \frac{w_1 - w_2}{w_1} = \frac{n}{n + N}$.
Here,$n$ is the number of moles of solute and $N$ is the number of moles of solvent (water).
$N = \frac{360 \ g}{18 \ g/mol} = 20 \ mol$.
Substituting the values: $\frac{3.5 - 2}{3.5} = \frac{1.5}{3.5} = \frac{3}{7} = \frac{n}{n + 20}$.
$3(n + 20) = 7n$ $\Rightarrow 3n + 60 = 7n$ $\Rightarrow 4n = 60$ $\Rightarrow n = 15 \ mol$.
Since $n = \frac{\text{mass of solute}}{\text{molar mass of solute}} = \frac{600}{M} = 15$.
$M = \frac{600}{15} = 40 \ g/mol$.

(C) The given equation for the vapour pressure of the solution is $P_S = 120 - 80 \, X_B$.
For pure liquid $A$,the mole fraction of $B$ is $X_B = 0$. Substituting this into the equation: $P_S = P_A^o = 120 - 80(0) = 120$.
For pure liquid $B$,the mole fraction of $B$ is $X_B = 1$. Substituting this into the equation: $P_S = P_B^o = 120 - 80(1) = 40$.
Thus,the vapour pressures of pure $A$ and $B$ are $120$ and $40$ respectively.

(C) For dilute solutions,the relative lowering of vapour pressure is equal to the mole fraction of the solute,which is approximately equal to the ratio of the number of moles of solute to the number of moles of solvent.
$\frac{P_0 - P_s}{P_0} \approx \frac{n_{\text{solute}}}{n_{\text{solvent}}}$
For the urea solution: $n_{\text{urea}} = \frac{1 \ g}{60 \ g/mol} = \frac{1}{60} \ mol$,$n_{\text{water}} = \frac{50 \ g}{18 \ g/mol} = \frac{50}{18} \ mol$.
Relative lowering for urea $= \frac{1/60}{50/18} = \frac{1}{60} \times \frac{18}{50} = \frac{18}{3000} = \frac{3}{500}$.
For the glucose solution: $n_{\text{glucose}} = \frac{w}{180} \ mol$,$n_{\text{water}} = \frac{100}{18} \ mol$.
Relative lowering for glucose $= \frac{w/180}{100/18} = \frac{w}{180} \times \frac{18}{100} = \frac{w}{1000}$.
Equating the two: $\frac{w}{1000} = \frac{3}{500} \implies w = \frac{3 \times 1000}{500} = 6 \ g$.

(A) The vapor pressure of pure solvent (water) is higher than the vapor pressure of the aqueous solution $(P_O > P_S)$.
Due to this difference in vapor pressure,water molecules evaporate from beaker $A$ and condense into beaker $B$ to reach equilibrium.
As a result,the volume of water in beaker $A$ decreases,and the volume of the solution in beaker $B$ increases.
Therefore,the level of water in beaker $A$ becomes lower than the level of the solution in beaker $B$.

(C) The total vapour pressure of a solution of two volatile liquids is given by $P_S = P_A^o X_A + P_B^o X_B$.
Since $X_A + X_B = 1$,we have $X_A = 1 - X_B$.
Substituting this into the equation: $P_S = P_A^o (1 - X_B) + P_B^o X_B = P_A^o + (P_B^o - P_A^o) X_B$.
Comparing this with the given equation $P_S = 150 - 60 X_B$:
For pure $A$ $(X_B = 0)$,$P_S = P_A^o = 150$.
For pure $B$ $(X_B = 1)$,$P_S = P_B^o = 150 - 60(1) = 90$.
Thus,the vapour pressures of pure $A$ and pure $B$ are $150$ and $90$ respectively.

(C) When a solution of $KCl$ in water is heated and the vapour (which is pure water) is removed,the amount of solvent decreases while the amount of solute $(KCl)$ remains constant.
This leads to an increase in the concentration of the solution (molality increases).
According to Raoult's Law,the vapour pressure of a solution is given by $P_{sol} = X_{solvent} \times P^0_{solvent}$.
As the concentration of the solute increases,the mole fraction of the solvent $(X_{solvent})$ decreases.
Consequently,the vapour pressure of the solution decreases as the solution becomes more concentrated.

(C) According to Raoult's Law,the total pressure $P$ of a binary mixture is given by $P = P_A^o X_A + P_B^o X_B$.
Since $X_A + X_B = 1$,we can write $X_B = 1 - X_A$.
Substituting this into the equation: $P = P_A^o X_A + P_B^o (1 - X_A) = P_B^o + (P_A^o - P_B^o) X_A$.
Comparing this with the given equation $P = 254 - 119 X_A$:
$P_B^o = 254 \ torr$.
$P_A^o - P_B^o = -119$.
$P_A^o - 254 = -119$.
$P_A^o = 254 - 119 = 135 \ torr$.
Thus,$P_A^o = 135 \ torr$ and $P_B^o = 254 \ torr$.