The vapor and liquid phases have identical compositions.

The boiling point is higher than that of either pure A or pure B.

The system deviates negatively from Raoult's Law.

The mole fraction of A in the vapor is always greater than that of B.

Shows a negative deviation from Raoult’s law

Shows no deviation from Raoult’s law

Shows positive deviation from Raoult’s law

Is saturated

${C_6}{H_6}\,\,and\,\,{C_6}{H_5}N{H_2}$

${C_6}{H_6}\,\,and\,\,{C_6}{H_5}I$

${C_6}{H_5}Cl\,\,and\,\,{C_2}{H_5}Br$

${C_6}{H_5}I\,\,and\,\,{C_6}{H_5}OH$

${P_A} + {X_A}\left( {{P_B} - {P_A}} \right)$

${P_A} + {X_A}\left( {{P_A} - {P_B}} \right)$

${P_B} + {X_A}\left( {{P_B} - {P_A}} \right)$

${P_B} + {X_A}\left( {{P_A} - {P_B}} \right)$

$76.0$

$752.4$

$759.0$

$7.6$

${P_A} + {X_A}\left( {{P_B} - {P_A}} \right)$

${P_A} + {X_A}\left( {{P_A} - {P_B}} \right)$

${P_B} + {X_A}\left( {{P_B} - {P_A}} \right)$

${P_B} + {X_A}\left( {{P_A} - {P_B}} \right)$

$\frac{{p - {p_s}}}{p} = \frac{n}{{n + N}}$

$\frac{{{p_s} - p}}{p} = \frac{N}{{N + n}}$

$\frac{{p - {p_s}}}{{{p_s}}} = \frac{N}{{N - n}}$

$\frac{{{p_s} - p}}{{{p_s}}} = \frac{{N - n}}{N}$

Benzene

Toluene

Both have equal partial pressures.

Cannot be determined.

Amount of liquid

Surface area of the container

Temperature

None of the above

440 torr

460torr

180 torr

88 torr

The relative lowering of vapour pressure is equal to mole fraction of solute

The lowering of vapour pressure is equal to the mole fraction of solute

The vapour pressure of the solution is equal to mole fraction of the solvent

The relative lowering of vapour pressure is proportional to amount of solute in solution