Lowering of vapour pressure Questions in English

(A) Given: $P^0 = 143\,mm \, Hg$,$w = 0.5\,g$,$m = 65\,g/mol$,$V = 100\,mL$,$d = 1.58\,g/cm^3$.
Mass of solvent $(W)$ $= V \times d = 100\,mL \times 1.58\,g/mL = 158\,g$.
Molar mass of $CCl_4$ $(M)$ $= 12 + 4 \times 35.5 = 154\,g/mol$.
Using Raoult's law: $\frac{P^0 - P_s}{P^0} = \frac{w \times M}{m \times W}$.
$\frac{143 - P_s}{143} = \frac{0.5 \times 154}{65 \times 158} = \frac{77}{10270} \approx 0.0075$.
$143 - P_s = 143 \times 0.0075 = 1.0725$.
$P_s = 143 - 1.0725 = 141.9275 \approx 141.93\,mm \, Hg$.

(B) $Raoult's$ law states that for any solution,the partial vapour pressure of each volatile component in the solution is directly proportional to its mole fraction.

(B) According to Raoult's law,for dilute solutions containing a non-volatile solute,the relative lowering of vapour pressure,$\frac{\Delta P}{P_0}$,is equal to the mole fraction of the solute,$X_2$.

(D) The relative lowering of vapour pressure is given by the formula: $\frac{P^0 - P_s}{P^0} = \frac{n}{n + N} \approx \frac{n}{N}$ (for dilute solutions).
Here,$n = \frac{w}{m} = \frac{71.5}{m}$ and $N = \frac{W}{M} = \frac{1000}{18} = 55.55 \ mol$.
Given $\frac{P^0 - P_s}{P^0} = 0.00713$.
Using the formula $\frac{P^0 - P_s}{P^0} = \frac{w \times M}{m \times W}$:
$0.00713 = \frac{71.5 \times 18}{m \times 1000}$
$m = \frac{71.5 \times 18}{0.00713 \times 1000} = \frac{1287}{7.13} \approx 180.5 \ g/mol$.
Thus,the molecular weight is approximately $180 \ g/mol$.

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

(A) According to Raoult's law,the partial vapour pressure of a volatile component in a solution is equal to the product of the vapour pressure of the pure component and its mole fraction in the solution.
For a solution containing a non-volatile solute,the vapour pressure of the solution $(P)$ is due only to the solvent.
Therefore,$P = P^o \times N_1$,where $N_1$ is the mole fraction of the solvent.
Since $N_1 + N_2 = 1$,we have $N_1 = 1 - N_2$.
Substituting this into the relative lowering of vapour pressure equation: $\frac{P^o - P}{P^o} = N_2$.
$1 - \frac{P}{P^o} = N_2 \implies \frac{P}{P^o} = 1 - N_2 = N_1$.
Thus,$P = P^o N_1$.
Hence,the correct option is $A$.

(C) Methanol $(CH_3OH)$ is more volatile than water $(H_2O)$ because it has a lower boiling point $(64.7 \ ^\circ C)$ compared to water $(100 \ ^\circ C)$.
According to Raoult's law,the vapour pressure of a solution containing a volatile solute is higher than the vapour pressure of the pure solvent if the solute is more volatile than the solvent.
Since methanol is more volatile than water,the vapour pressure of the aqueous solution of methanol will be higher than that of pure water.

(A) The vapour pressure of a solution decreases as the number of solute particles increases (colligative property).
$1 \ M \ C_{12}H_{22}O_{11}$ (sucrose) is a non-electrolyte and produces $1$ particle per molecule.
$1 \ M \ CH_3COOH$ is a weak electrolyte and produces slightly more than $1$ particle.
$1 \ M \ NaCl$ dissociates into $2$ particles ($Na^+$ and $Cl^-$).
$1 \ M \ CaCl_2$ dissociates into $3$ particles ($Ca^{2+}$ and $2Cl^-$).
Since sucrose has the minimum number of particles,it will have the minimum lowering of vapour pressure,and therefore,the maximum vapour pressure.

(B) According to Raoult's law for a dilute solution containing a non-volatile solute,the relative lowering of vapour pressure is given by the expression: $\frac{P^o - P}{P^o} = X_{solute} = \frac{n}{n + N}$.
Here,$n$ is the number of moles of solute and $N$ is the number of moles of solvent.
Therefore,the relative lowering of vapour pressure is equal to the ratio of the number of solute molecules to the total number of molecules in the solution $(n + N)$.

(B) According to Raoult's law,when a non-volatile solute is added to a solvent,the vapour pressure of the solution decreases.
Even though acetone is volatile,the addition of a solute to a solvent generally lowers the chemical potential of the solvent,leading to a decrease in the vapour pressure of the solvent component in the mixture compared to its pure state.
Therefore,the vapour pressure of water over the solution will be less than the vapour pressure of pure water.

(C) According to the relative lowering of vapour pressure formula:
$\frac{P^{\circ} - P}{P^{\circ}} = X_2$
Where:
$P^{\circ} = 50 \ mm$ (vapour pressure of pure solvent)
$P = 45 \ mm$ (vapour pressure of solution)
$X_2$ = mole fraction of the solute
Substituting the values:
$\frac{50 - 45}{50} = X_2$
$\frac{5}{50} = X_2$
$X_2 = 0.1$

(B) Given molecular mass of sucrose $= 342 \, g/mol$.
Moles of sucrose $(n) = \frac{100}{342} \approx 0.292 \, mol$.
Moles of water $(N) = \frac{1000}{18} \approx 55.56 \, mol$.
Vapour pressure of pure water $(P^0) = 23.8 \, mm \, Hg$.
According to Raoult's law,the relative lowering of vapour pressure is given by:
$\frac{\Delta P}{P^0} = \frac{n}{n + N}$
Substituting the values:
$\frac{\Delta P}{23.8} = \frac{0.292}{0.292 + 55.56} = \frac{0.292}{55.852} \approx 0.005228$
$\Delta P = 23.8 \times 0.005228 \approx 0.1244 \, mm \, Hg \approx 0.125 \, mm \, Hg$.

(D) and $D$ are incorrect. However,$D$ is the most fundamentally incorrect statement regarding the definition of relative lowering of vapour pressure. According to Raoult's law,the relative lowering of vapour pressure is given by $\frac{P^o - P}{P^o} = X_{solute}$. This value depends only on the mole fraction of the solute and is independent of the nature of the solute or the original vapour pressure of the solvent. Furthermore,vapour pressure itself is not a colligative property; the relative lowering of vapour pressure is the colligative property.

(B) Vapour pressure is inversely proportional to the strength of intermolecular forces of attraction.
Mercury $(Hg)$ is a metal with strong metallic bonding between its atoms.
In contrast,water,kerosene,and rectified spirit are held together by weaker intermolecular forces like hydrogen bonding or van der Waals forces.
Due to the very strong metallic bonds in mercury,its tendency to escape into the vapour phase is minimal at room temperature.
Therefore,mercury exerts the lowest vapour pressure among the given substances.

(B) According to Raoult's law,for a solution containing a non-volatile solute,the relative lowering of vapour pressure is given by the expression: $\frac{P^{\circ} - P_s}{P^{\circ}} = X_2$.
Here,$\frac{P^{\circ} - P_s}{P^{\circ}}$ represents the relative lowering of vapour pressure,and $X_2$ is the mole fraction of the solute.
Thus,the relative lowering of vapour pressure is equal to the mole fraction of the solute.

(A) The vapour pressure of a liquid is a characteristic property that depends on the nature of the liquid and the temperature. As the temperature increases,the kinetic energy of the molecules increases,leading to a higher rate of evaporation and thus higher vapour pressure. It is independent of the volume of the liquid or the container,provided the liquid is in equilibrium with its vapour.

(B) According to Raoult's Law for a non-volatile solute:
$\frac{P^0 - P_s}{P^0} = x_B$
Where $P^0$ is the vapour pressure of pure solvent $(0.80 \ atm)$,
$P_s$ is the vapour pressure of the solution $(0.6 \ atm)$,
and $x_B$ is the mole fraction of the solute $B$.
Substituting the values:
$x_B = \frac{0.80 - 0.60}{0.80} = \frac{0.20}{0.80} = 0.25$.

(D) The relative lowering of vapour pressure is a colligative property,which depends on the van't Hoff factor $(i)$.
For a solution,$\frac{P^0 - P_s}{P^0} = i \times \text{mole fraction of solute}$.
Since the concentration is the same for all,the lowering of vapour pressure is proportional to the van't Hoff factor $(i)$.
$(A)$ Urea: $i = 1$
$(B)$ Glucose: $i = 1$
$(C)$ $MgSO_4$: $i = 2$ (dissociates into $Mg^{2+}$ and $SO_4^{2-}$)
$(D)$ $BaCl_2$: $i = 3$ (dissociates into $Ba^{2+}$ and $2Cl^-$)
Since $BaCl_2$ has the highest van't Hoff factor $(i = 3)$,it shows the highest lowering of vapour pressure.

(A) The relative lowering in vapour pressure is given by the formula: $\frac{P^0 - P_s}{P^0} = \frac{n_2}{n_1 + n_2}$.
Here,the number of moles of glucose $(n_2) = \frac{18 \ g}{180 \ g/mol} = 0.1 \ mol$.
The number of moles of water $(n_1) = \frac{90 \ g}{18 \ g/mol} = 5 \ mol$.
Since $n_2 << n_1$,we can approximate the formula as $\frac{n_2}{n_1} = \frac{0.1}{5} = 0.02$.
Thus,the relative lowering in vapour pressure is $0.02$.

(C) The given statement defines Raoult's law for solutions containing non-volatile solutes.
According to Raoult's law,the relative lowering of vapour pressure is given by the expression: $\frac{P^0 - P}{P^0} = X_{solute}$
Where:
$P^0$ = Vapour pressure of pure solvent
$P$ = Vapour pressure of the solution
$X_{solute}$ = Mole fraction of the solute
This shows that the relative lowering of vapour pressure is directly proportional to the mole fraction of the non-volatile solute.

(D) According to Raoult's law,the total vapour pressure $P_T$ of an ideal solution is given by:
$P_T = P_P^0 X_P + P_Q^0 X_Q$
Where $P_P^0 = 80 \, torr$ and $P_Q^0 = 60 \, torr$.
The mole fractions are:
$X_P = \frac{3}{3+2} = \frac{3}{5} = 0.6$
$X_Q = \frac{2}{3+2} = \frac{2}{5} = 0.4$
Substituting the values:
$P_T = (80 \times 0.6) + (60 \times 0.4)$
$P_T = 48 + 24 = 72 \, torr$.

(C) According to Raoult's law for a non-volatile solute,the relative lowering of vapour pressure is given by:
$\frac{P^0 - P_s}{P^0} = \frac{n}{n + N} \approx \frac{n}{N} = \frac{w \times M}{m \times W}$
Where:
$P^0 = 640 \, mm \, Hg$ (Vapour pressure of pure benzene)
$P_s = 600 \, mm \, Hg$ (Vapour pressure of solution)
$w = 2.175 \, g$ (Mass of solute)
$W = 39.08 \, g$ (Mass of benzene)
$M = 78 \, g/mol$ (Molar mass of benzene,$C_6H_6$)
$m$ = Molar mass of solute
Substituting the values:
$\frac{640 - 600}{640} = \frac{2.175 \times 78}{m \times 39.08}$
$\frac{40}{640} = \frac{169.65}{m \times 39.08}$
$m = \frac{169.65 \times 640}{39.08 \times 40}$
$m = 69.45 \, g/mol$
Thus,the molecular weight of the solid substance is approximately $69.5 \, g/mol$.

(A) According to Raoult's law,the relative lowering of vapour pressure is equal to the mole fraction of the solute.
$\frac{p - p_s}{p} = X_{solute}$
Since the mole fraction of the solute $X_{solute}$ is given by $\frac{n}{n + N}$,where $n$ is the number of moles of the solute and $N$ is the number of moles of the solvent.
Therefore,the expression is $\frac{p - p_s}{p} = \frac{n}{n + N}$.
Thus,the correct answer is $A$.

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

(C) $1$. Calculate the moles of solute (Urea): $n = \frac{\text{mass}}{\text{molar mass}} = \frac{60 \ g}{60 \ g/mol} = 1 \ mol$.
$2$. Moles of solvent (water) is given as $N = 9.9 \ mol$.
$3$. According to Raoult's Law for non-volatile solutes,the relative lowering of vapour pressure is given by: $\frac{P_o - P_s}{P_o} = \frac{n}{n + N}$.
$4$. Substituting the values: $\frac{P_o - P_s}{P_o} = \frac{1}{1 + 9.9} = \frac{1}{10.9} \approx 0.0917$.
$5$. Therefore,$1 - \frac{P_s}{P_o} = 0.0917 \Rightarrow \frac{P_s}{P_o} = 1 - 0.0917 = 0.9083$.
$6$. Thus,$P_s \approx 0.90 \ P_o$.

(B) Given: $P^{\circ} = 17.54\,mm$,$\Delta P = 0.30\,mm$,$w = 20\,g$,$W = 100\,g$,$M = 18\,g/mol$.
The relative lowering of vapour pressure is given by the formula: $\frac{P^{\circ} - P_S}{P^{\circ}} = \frac{w \times M}{m \times W}$.
Alternatively,using the relation $\frac{\Delta P}{P_S} = \frac{w \times M}{m \times W}$ where $P_S = P^{\circ} - \Delta P = 17.54 - 0.30 = 17.24\,mm$.
Substituting the values: $\frac{0.30}{17.24} = \frac{20 \times 18}{m \times 100}$.
$m = \frac{20 \times 18 \times 17.24}{0.30 \times 100} = \frac{360 \times 17.24}{30} = 12 \times 17.24 = 206.88\,g/mol$.

(C) According to Raoult's Law for non-volatile solutes: $\frac{P^o - P_s}{P^o} = \frac{n_2}{n_1 + n_2} \approx \frac{n_2}{n_1} = \frac{w_2 \times M_1}{M_2 \times w_1}$.
Given: $P^o = 195 \ mm \ Hg$,$P_s = 192.5 \ mm \ Hg$,$w_2 = 1 \ g$,$w_1 = 100 \ g$,$M_1 = 58 \ g/mol$.
Substituting the values: $\frac{195 - 192.5}{195} = \frac{1 \times 58}{M_2 \times 100}$.
$\frac{2.5}{195} = \frac{58}{100 \times M_2}$.
$M_2 = \frac{58 \times 195}{2.5 \times 100} = \frac{11310}{250} = 45.24 \ g/mol$.

(B) 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 present in the solution.
Mathematically,it is expressed as: $\frac{P^o - P_s}{P^o} = x_2$,where $P^o$ is the vapour pressure of the pure solvent,$P_s$ is the vapour pressure of the solution,and $x_2$ is the mole fraction of the solute.

(A) Given: The vapour pressure of pure solvent $A$ is $P_0 = 0.80 \, atm$.
After adding the non-volatile solute $B$,the vapour pressure of the solution is $P = 0.6 \, atm$.
According to Raoult's Law for the relative lowering of vapour pressure:
$\frac{P_0 - P}{P_0} = X_B$
Substituting the values:
$\frac{0.80 - 0.6}{0.80} = X_B$
$\frac{0.2}{0.8} = X_B$
$X_B = 0.25$
Thus,the mole fraction of $B$ in the solution is $0.25$.

(D) According to Raoult's Law,the mole fraction of a component in the vapour phase $(y_A)$ is given by the formula:
$y_A = \frac{P_A}{P_{total}} = \frac{x_A \cdot P_A^0}{x_A \cdot P_A^0 + x_B \cdot P_B^0}$
Given:
$P_A^0 : P_B^0 = 1 : 2 \implies P_A^0 = 1, P_B^0 = 2$
$x_A : x_B = 1 : 2 \implies x_A = 1, x_B = 2$
Substituting these values into the formula:
$y_A = \frac{1 \times 1}{(1 \times 1) + (2 \times 2)}$
$y_A = \frac{1}{1 + 4} = \frac{1}{5} = 0.2$
Therefore,the mole fraction of $A$ in the vapour phase is $0.2$.

(D)
The lowering of vapour pressure is a colligative property,which depends on the number of solute particles (van't Hoff factor,$i$) in the solution.
$1.$ $NaCl \to Na^+ + Cl^-$ $(i = 2)$
$2.$ $\text{Sucrose} \to \text{Sucrose}$ $(i = 1)$
$3.$ $BaCl_2 \to Ba^{2+} + 2Cl^-$ $(i = 3)$
$4.$ $Na_3PO_4 \to 3Na^+ + PO_4^{3-}$ $(i = 4)$
Since $Na_3PO_4$ produces the highest number of ions $(i = 4)$,it will cause the greatest lowering of vapour pressure,resulting in the lowest vapour pressure among the given solutions.

(C) The vapour pressure of a solvent is lowered by the presence of a solute.
Lowering in vapour pressure is a colligative property,which depends on the number of particles (ions) present in the solution.
$0.1 \ M$ sugar solution is a non-electrolyte and provides $1$ particle.
$0.1 \ M \ KCl$ dissociates into $K^+$ and $Cl^-$,providing $2$ ions.
$0.1 \ M \ AgNO_3$ dissociates into $Ag^+$ and $NO_3^-$,providing $2$ ions.
$0.1 \ M \ Cu(NO_3)_2$ dissociates into $Cu^{2+}$ and $2NO_3^-$,providing $3$ ions.
Since $Cu(NO_3)_2$ produces the maximum number of ions,it causes the greatest lowering in vapour pressure.
Therefore,the vapour pressure is lowest for $0.1 \ M \ Cu(NO_3)_2$ solution.

(D) The vapour pressure of a solution decreases as the number of solute particles (van't Hoff factor,$i$) increases.
$1$. $0.1 \ M \ KCl$ dissociates into $2$ ions $(K^+, Cl^-)$,so $i = 2$.
$2$. $0.1 \ M$ urea is a non-electrolyte,so $i = 1$.
$3$. $0.1 \ M \ Na_2SO_4$ dissociates into $3$ ions $(2Na^+, SO_4^{2-})$,so $i = 3$.
$4$. $0.1 \ M \ K_4[Fe(CN)_6]$ dissociates into $5$ ions $(4K^+, [Fe(CN)_6]^{4-})$,so $i = 5$.
Since $K_4[Fe(CN)_6]$ produces the maximum number of ions,it results in the greatest lowering of vapour pressure,meaning it has the lowest vapour pressure.

(A) According to Ostwald-Walker method,the relative lowering of vapour pressure is given by: $\frac{p^o - p}{p^o} = \frac{w \times M}{W \times m}$.
Here,loss in weight of solution is proportional to $p$ and loss in weight of solvent is proportional to $(p^o - p)$.
Given: weight of solute $(w) = 5 \ g$,weight of solvent $(W) = 80 \ g$,loss in solution $= 2.50 \ g$,loss in solvent $= 0.04 \ g$.
Formula: $\frac{p^o - p}{p} = \frac{\text{loss in solvent}}{\text{loss in solution}} = \frac{0.04}{2.50}$.
Also,$\frac{p^o - p}{p} = \frac{n}{N} = \frac{w/m}{W/M} = \frac{w \times M}{W \times m}$.
Substituting values: $\frac{0.04}{2.50} = \frac{5 \times 18}{80 \times m}$.
$m = \frac{5 \times 18 \times 2.50}{80 \times 0.04} = \frac{225}{3.2} = 70.31 \ g/mol$.

(A) According to Raoult's law,the relative lowering of vapour pressure is equal to the mole fraction of the solute.
The formula is: $\frac{P^o - P_s}{P^o} = X_{solute}$
Given: $P^o = 0.80 \ atm$,$P_s = 0.60 \ atm$
Substituting the values: $X_{solute} = \frac{0.80 - 0.60}{0.80} = \frac{0.20}{0.80} = 0.25$
Therefore,the mole fraction of the solute is $0.25$.

(C) Let $M$ be the molecular weight of the solute and $P^0$ be the vapour pressure of pure water at $25 \, ^oC$.
According to Raoult's law for dilute solutions,$\frac{P^0 - P_s}{P_s} = \frac{n_2}{n_1} = \frac{w_2 \times M_1}{M_2 \times w_1}$.
For the first case: $\frac{P^0 - 21.85}{21.85} = \frac{30 \times 18}{M \times 90} = \frac{6}{M} \implies P^0 - 21.85 = \frac{131.1}{M} \dots (i)$.
For the second case,after adding $18 \, g$ of water,the new mass of water is $90 + 18 = 108 \, g$: $\frac{P^0 - 22.15}{22.15} = \frac{30 \times 18}{M \times 108} = \frac{5}{M} \implies P^0 - 22.15 = \frac{110.75}{M} \dots (ii)$.
Subtracting equation $(ii)$ from $(i)$: $(P^0 - 21.85) - (P^0 - 22.15) = \frac{131.1}{M} - \frac{110.75}{M}$.
$0.30 = \frac{20.35}{M}$.
$M = \frac{20.35}{0.30} = 67.83 \, g \, mol^{-1}$.

(C) According to Raoult's law for dilute solutions,the relative lowering of vapour pressure is equal to the mole fraction of the solute: $\frac{P^o - P_s}{P^o} = \frac{n_2}{n_1} = \frac{W_2 / M_2}{W_1 / M_1}$.
Given: $P^o = 3000 \ N/m^2$,$P_s = 2985 \ N/m^2$,$W_2 = 5 \ g$,$W_1 = 100 \ g$,$M_1 = 18 \ g/mol$ (for water).
Substituting the values: $\frac{3000 - 2985}{3000} = \frac{5 / M_2}{100 / 18}$.
$\frac{15}{3000} = \frac{5 \times 18}{100 \times M_2}$.
$0.005 = \frac{90}{100 \times M_2}$.
$0.005 = \frac{0.9}{M_2}$.
$M_2 = \frac{0.9}{0.005} = 180 \ g/mol$.

(A) The relative lowering of vapour pressure is given by $\frac{P^o - P}{P^o} = \chi_{solute} = \frac{n}{n + N}$.
Since the amount of solvent (water) is the same $(1 \ L)$ and the amount of solute is small,the mole fraction is directly proportional to the number of moles of solute $(n)$.
For the first solution,moles of urea $= \frac{12 \ g}{60 \ g/mol} = 0.2 \ mol$.
For the second solution,moles of cane sugar $= \frac{68.4 \ g}{342 \ g/mol} = 0.2 \ mol$.
Since the number of moles of solute is the same in both solutions,the lowering of vapour pressure will be the same.

(C) Methanol is more volatile than water.
When methanol is dissolved in water,the resulting solution exhibits a positive deviation from Raoult's law.
This occurs because the intermolecular forces between $CH_3OH$ and $H_2O$ are weaker than the forces in pure $CH_3OH$ or pure $H_2O$.
Consequently,the vapour pressure of the solution is higher than that of pure water.

(D) The relative lowering of vapour pressure is given by the formula: $\frac{P^\circ - P_s}{P^\circ} = \frac{n_2}{n_1 + n_2} \approx \frac{n_2}{n_1}$ (for dilute solutions).
Here,$P^\circ = 17.5 \ mm \ Hg$,$W_{solute} = 18 \ g$,$M_{solute} = 180 \ g/mol$,$W_{solvent} = 178.2 \ g$,$M_{solvent} = 18 \ g/mol$.
Moles of glucose $(n_2) = \frac{18}{180} = 0.1 \ mol$.
Moles of water $(n_1) = \frac{178.2}{18} = 9.9 \ mol$.
Using the formula $\frac{P^\circ - P_s}{P^\circ} = \frac{n_2}{n_1 + n_2}$:
$\frac{17.5 - P_s}{17.5} = \frac{0.1}{9.9 + 0.1} = \frac{0.1}{10} = 0.01$.
$17.5 - P_s = 17.5 \times 0.01 = 0.175$.
$P_s = 17.5 - 0.175 = 17.325 \ mm \ Hg$.

(D) Given: $P_A^0 = 150 \ mm \ Hg$,$P_B^0 = 100 \ mm \ Hg$.
Since the mole fractions are equal,$X_A = X_B = 0.5$.
Total pressure $P_{total} = P_A^0 X_A + P_B^0 X_B = (150 \times 0.5) + (100 \times 0.5) = 75 + 50 = 125 \ mm \ Hg$.
According to Dalton's Law,the partial pressure of $B$ in the vapor phase is $P_B = P_B^0 X_B = 100 \times 0.5 = 50 \ mm \ Hg$.
The mole fraction of $B$ in the vapor phase $(Y_B)$ is given by $Y_B = \frac{P_B}{P_{total}} = \frac{50}{125} = 0.4$.

(A) According to Raoult's law,the total vapor pressure $P_s = P_A^0 X_A + P_B^0 X_B$.
For the first solution: $X_A = 1/3$ and $X_B = 2/3$. Thus,$250 = P_A^0(1/3) + P_B^0(2/3)$,which simplifies to $P_A^0 + 2P_B^0 = 750$ (Equation $1$).
For the second solution: $1 \ mol$ of $A$ is added,so $n_A = 2 \ mol$ and $n_B = 2 \ mol$. Thus,$X_A = 2/4 = 0.5$ and $X_B = 2/4 = 0.5$. The total pressure is $300 = P_A^0(0.5) + P_B^0(0.5)$,which simplifies to $P_A^0 + P_B^0 = 600$ (Equation $2$).
Subtracting Equation $2$ from Equation $1$: $(P_A^0 + 2P_B^0) - (P_A^0 + P_B^0) = 750 - 600$,giving $P_B^0 = 150 \ mm \ Hg$.
Substituting $P_B^0$ into Equation $2$: $P_A^0 + 150 = 600$,so $P_A^0 = 450 \ mm \ Hg$.

(B) According to Raoult's Law for solutions containing non-volatile solutes,the relative lowering of vapour pressure is equal to the mole fraction of the solute.
$\frac{P_0 - P_s}{P_0} = N_2$
Rearranging the equation:
$P_0 - P_s = P_0N_2$

(A) The relative lowering of vapour pressure is given by the formula: $\frac{P^\circ - P_s}{P_s} = \frac{n}{N} = \frac{w \times M}{m \times W}$
Given:
$m = 40 \ g/mol$ (molar mass of solute)
$W = 114 \ g$ (mass of octane)
$M = 114 \ g/mol$ (molar mass of octane,$C_8H_{18}$)
If vapour pressure decreases by $20\%$,then $P_s = 0.8 P^\circ$.
Substituting the values:
$\frac{P^\circ - 0.8 P^\circ}{0.8 P^\circ} = \frac{w \times 114}{40 \times 114}$
$\frac{0.2}{0.8} = \frac{w}{40}$
$0.25 = \frac{w}{40}$
$w = 0.25 \times 40 = 10 \ g$

(B) The relative lowering of vapour pressure is given by the formula: $\frac{p^0 - p}{p^0} = \frac{n_2}{n_1 + n_2} \approx \frac{n_2}{n_1} = \frac{W \times M_0}{M \times W_0}$.
For urea $(M = 60 \ g/mol)$: $\frac{p^0 - p}{p^0} = \frac{1 \times 18}{60 \times 50}$.
For glucose $(M = 180 \ g/mol)$: $\frac{p^0 - p}{p^0} = \frac{W \times 18}{180 \times 50}$.
Since the relative lowering of vapour pressure is equal for both:
$\frac{1 \times 18}{60 \times 50} = \frac{W \times 18}{180 \times 50}$.
Solving for $W$:
$\frac{1}{60} = \frac{W}{180} \implies W = \frac{180}{60} = 3 \ g$.

(C) According to Raoult's law for an ideal solution,the total vapor pressure $P_s$ is given by:
$P_s = P_A^0 X_A + P_B^0 X_B$
Here,$n_{heptane} = 1 \ mol$,$n_{octane} = 4 \ mol$.
Total moles = $1 + 4 = 5 \ mol$.
Mole fraction of heptane $(X_A)$ = $1/5 = 0.2$.
Mole fraction of octane $(X_B)$ = $4/5 = 0.8$.
$P_A^0 = 92 \ mm \ Hg$,$P_B^0 = 31 \ mm \ Hg$.
$P_s = (92 \times 0.2) + (31 \times 0.8)$
$P_s = 18.4 + 24.8 = 43.2 \ mm \ Hg$.

(A) According to Raoult's Law for a non-volatile solute,the relative lowering of vapor pressure is given by: $\frac{P^o_A - P_s}{P^o_A} = \frac{n_B}{n_A + n_B} \approx \frac{n_B}{n_A}$.
Given: $P^o_A = 10 \ torr$,$P_s = 9 \ torr$,$w_B = 1 \ g$,$w_A = 20 \ g$,$M_A = 200 \ g/mol$.
Substituting the values: $\frac{10 - 9}{10} = \frac{w_B / M_B}{w_A / M_A}$.
$0.1 = \frac{1 / M_B}{20 / 200}$.
$0.1 = \frac{1 / M_B}{0.1}$.
$0.1 \times 0.1 = \frac{1}{M_B}$.
$0.01 = \frac{1}{M_B}$.
$M_B = \frac{1}{0.01} = 100 \ g/mol$.

(C) According to Raoult's Law,the total pressure of the solution $(P_{total})$ is given by:
$P_{total} = P_A^0 \cdot X_A + P_B^0 \cdot X_B$
Given:
$P_A^0 = 100 \ Torr$,$P_B^0 = 80 \ Torr$
$n_A = 2 \ moles$,$n_B = 3 \ moles$
Total moles $= 2 + 3 = 5 \ moles$
Mole fraction of $A$ $(X_A) = \frac{2}{5} = 0.4$
Mole fraction of $B$ $(X_B) = \frac{3}{5} = 0.6$
$P_{total} = (100 \times 0.4) + (80 \times 0.6)$
$P_{total} = 40 + 48 = 88 \ Torr$

(C) The loss in weight of the solution is proportional to the vapour pressure of the solution $(p)$.
The loss in weight of the solvent is proportional to the lowering of vapour pressure $(p^0 - p)$.
According to the Ostwald-Walker method,the relative lowering of vapour pressure is given by:
$\frac{p^0 - p}{p} = \frac{\text{Loss in weight of solvent}}{\text{Loss in weight of solution}}$
Also,$\frac{p^0 - p}{p} = \frac{w \times M_0}{m \times W_0}$,where $w$ is the mass of solute,$m$ is the molar mass of solute,$W_0$ is the mass of solvent,and $M_0$ is the molar mass of water $(18 \ g/mol)$.
Substituting the values:
$\frac{0.05}{2.5} = \frac{10 \times 18}{m \times 90}$
$0.02 = \frac{180}{90m}$
$0.02 = \frac{2}{m}$
$m = \frac{2}{0.02} = 100 \ g/mol$.

(C) According to Raoult's law for a solution containing a non-volatile solute,the relative lowering of vapour pressure is given by: $\frac{P^\circ - P_s}{P^\circ} = \frac{n}{n + N}$.
From this,we can derive other forms:
$1$. $\frac{P_s}{P^\circ} = 1 - \frac{n}{n + N} = \frac{N}{n + N}$ (Correct,Option $A$)
$2$. $\frac{P^\circ - P_s}{P_s} = \frac{n}{N}$ (Correct,Option $C$ is $\frac{n}{n+N}$,which is incorrect as it should be $\frac{n}{N}$)
$3$. $\frac{P^\circ}{P^\circ - P_s} = \frac{n + N}{n} = 1 + \frac{N}{n}$ (Correct,Option $B$)
$4$. $\frac{P_s}{P^\circ - P_s} = \frac{N}{n}$ (Correct,Option $D$)
Option $C$ is the incorrect form because the denominator on the left side should be $P_s$ and the right side should be $\frac{n}{N}$ for the ratio to be correct,or the expression $\frac{P^\circ - P_s}{P_s} = \frac{n}{n+N}$ is mathematically inconsistent with the standard definition.