Solutions

A binary mixture of $\text{C}_6\text{H}_5\text{OH}$ and $\text{C}_6\text{H}_5\text{NH}_2$ exhibits negative deviation from Raoult's law.

This means that the vapor pressure of the solution is lower than the vapor pressures of the pure components, $\text{C}_6\text{H}_5\text{OH}$ and $\text{C}_6\text{H}_5\text{NH}_2$.

Consequently, the boiling point of the solution is higher than the boiling points of the pure substances.

Therefore, this mixture forms a maximum boiling azeotrope.

th, .header { background-color: #0000FF; color: white; } Salt Values of i (for different conc. of a Salt) NaCl 0.1 M 0.01 M 0.001 M 1.87 1.94 1.94 The van 't Hoff factor (i) is used to describe the number of particles a solute formula unit produces in a solution.

For an electrolyte like $\mathrm{NaCl}$, which dissociates completely in very dilute solutions, the theoretical value of $i$ is approximately 2, since $\mathrm{NaCl}$ dissociates into $\mathrm{Na}^+$ and $\mathrm{Cl}^-$ ions.In real scenarios, as the concentration of the solution decreases (making the solution more dilute), the interaction between the ions decreases, allowing more complete dissociation.

Therefore, for practical purposes, the van 't Hoff factor $i$ approaches its theoretical maximum value as concentration decreases.

Thus, for $\mathrm{NaCl}$ solutions of concentrations $0.1 \mathrm{M}$, $0.01 \mathrm{M}$, and $0.001 \mathrm{M}$, the fact that $\mathrm{NaCl}$ dissociates more completely in more dilute solutions implies that $i$ increases with decreasing concentration.Hence, the order of $i$ based on the concentration would be $\mathrm{i}_{\mathrm{A}} 0.01 \mathrm{M} > 0.001 \mathrm{M}$, respectively.

Thus, option B correctly describes the order of the van 't Hoff factors for these solutions.

Given: 1 mole of volatile liquid A 3 moles of volatile liquid B Vapor pressure of pure A, $P_A^o = 200$ mm Hg Vapor pressure of the solution, $P_{S} = 500$ mm Hg We apply Raoult's law, which states: $P_{S} = P_A^o \cdot X_A + P_B^o \cdot X_B$ Where: $X_A$ is the mole fraction of A $X_B$ is the mole fraction of B $P_B^o$ is the vapor pressure of pure liquid B Calculate the mole fractions: $X_A = \dfrac{1}{1+3} = \dfrac{1}{4}$ $X_B = \dfrac{3}{1+3} = \dfrac{3}{4}$ Plug these into the equation: $500 = 200 \times \dfrac{1}{4} + P_B^o \times \dfrac{3}{4}$ Simplifying: $500 = 50 + \dfrac{3}{4} P_B^o$ Subtract 50 from both sides: $450 = \dfrac{3}{4} P_B^o$ Multiply both sides by $\dfrac{4}{3}$ to solve for $P_B^o$: $P_B^o = 600 \, \text{mm Hg}$ Since $P_A^o In conclusion: The vapor pressure of pure B,$ P_B^o $, is 600 mm Hg.

The least volatile component is A.

Gaseous densities of ethanol and dimethyl ether would be same at same temperature and pressure.

The heat of vaporisation ,

and

will differ due to

-bonding in ethanol.