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A system of greater disorder of molecule...

A system of greater disorder of molecules is more probable. The disorder of molecules is reflected by the entropy of the system.A liquid vapourises to form a more disordered gas.When a solute is present, there is additional contribution to the entropy of the liquid due to increased randomness.As the entropy of solution is higher than that of pure liquid, there is weaker tendency to form the gas.Thus, a solute (non-volatile) lowers the vapour pressure of a liquid, and hence a higher boiling point of the solution.
Similarly, the greater randomness of the solution opposes the tendency to freeze.In consequence, a lower the temperature must be reached for achieving the equilibrium between the solid (frozen solvent) and the solution.Elevation of `B.Pt.(DeltaT_b)` and depression of `F.Pt.(DeltaT_f)` of a solution are the colligative properties which depend only on the concentration of particles of the solute, not their identify.For dilute solutions. `DeltaT_b and DeltaT_f` are proportional to the molality of the solute in the solution.
`DeltaT_b=K_bm , K_b`=Ebullioscopic constant=`(RT_(b)^(@^(2))M)/(1000 DeltaH_(vap))`
And `DeltaT_f=K_fm , K_f`=Cryoscopic constant=`(RT_(f)^(@^(2))M)/(1000 DeltaH_(fus))` (M=molecular mass of the sovent)
The values of `K_b and K_f` do depend on the properties of the solvent.For liquids, `(DeltaH_(vap))/T_b^@` is almost constant.
[Troutan's Rule, this constant for most of the unassociated liquids (not having any strong bonding like Hydrogen bonding in the liquid state ) is equal to `90 J//"mol"`.]
For solutes undergoing changes of molecular state is solution (ionization or association), the observed `DeltaT` values differ from the calculated ones using the above relations In such situations, the relationships are modified as `DeltaT_b=i K_bm, DeltaT_f=i K_fm`
where i=Van't Hoff factor, greater than unity for ionization and smaller than unity for association of the solute molecules.
A mixture of two Immiscible liquids at a constant pressure of 1 atm boils at a temperature

A

equal to the normal boiling point of more volatile liquid

B

equal to the mean of the normal boiling point of the two liquids

C

greater than the normal boiling point of either of the liquid

D

smaller than the normal boiling point of either of the liquid

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To solve the question regarding the boiling point of a mixture of two immiscible liquids at a constant pressure of 1 atm, we can follow these steps: ### Step 1: Understand the Concept of Boiling Point The boiling point of a liquid is defined as the temperature at which its vapor pressure equals the external atmospheric pressure. In this case, the atmospheric pressure is given as 1 atm. ### Step 2: Analyze the Properties of Immiscible Liquids When two immiscible liquids are mixed, the total vapor pressure of the mixture is the sum of the vapor pressures of the individual liquids. This can be expressed as: \[ P_{\text{mixture}} = P_1 + P_2 \] where \( P_1 \) is the vapor pressure of liquid 1 and \( P_2 \) is the vapor pressure of liquid 2. ### Step 3: Determine the Effect on Vapor Pressure Since both \( P_1 \) and \( P_2 \) are positive values, the vapor pressure of the mixture \( P_{\text{mixture}} \) will be greater than either \( P_1 \) or \( P_2 \): \[ P_{\text{mixture}} > P_1 \quad \text{and} \quad P_{\text{mixture}} > P_2 \] ### Step 4: Relate Vapor Pressure to Boiling Point As the vapor pressure of the mixture increases, the boiling point decreases. This is because a higher vapor pressure means that the liquid will reach the atmospheric pressure at a lower temperature. ### Step 5: Conclusion on Boiling Point Since the vapor pressure of the mixture is greater than that of either liquid, the boiling point of the mixture will be lower than the boiling points of both individual liquids. Therefore, we conclude that: \[ \text{Boiling point of the mixture} < \text{Boiling point of liquid 1} \quad \text{and} \quad \text{Boiling point of the mixture} < \text{Boiling point of liquid 2} \] ### Final Answer The correct option is: **Smaller than the normal boiling point of either of the liquids.** ---

To solve the question regarding the boiling point of a mixture of two immiscible liquids at a constant pressure of 1 atm, we can follow these steps: ### Step 1: Understand the Concept of Boiling Point The boiling point of a liquid is defined as the temperature at which its vapor pressure equals the external atmospheric pressure. In this case, the atmospheric pressure is given as 1 atm. ### Step 2: Analyze the Properties of Immiscible Liquids When two immiscible liquids are mixed, the total vapor pressure of the mixture is the sum of the vapor pressures of the individual liquids. This can be expressed as: \[ P_{\text{mixture}} = P_1 + P_2 \] ...
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A system of greater disorder of molecules is more probable. The disorder of molecules is reflected by the entropy of the system. A liquid vapourizes to form a more disordered gas. When a solute is present, there is additional contribution to the entropy of the liquid due to increased randomness. As the entropy of solution is higher than that of pure liquid, there is weaker tendency to form the gas. Thus, a solute (non-volatile) lowers the vapour pressure of a liquid, and hence a higher boiling point of the solution. Similarly, the greater randomness of the solution opposes the tendercy to freeze. In consequence, a lower temperature must be reached for achieving the equilibrium between the solid (frozen solvent) and the solution. The elevation in boiling point (DeltaT_(b)) and depression in freezing point (DeltaT_(f)) of a solution are the colligative properties which depend only on the concentration of particles of the solute and not their identity. For dilute solutions, (DeltaT_(b)) and (DeltaT_(f)) are proportional to the molarity of the solute in the solution. Dissolution of a non-volatile solute into a liquid leads to

A system of greater disorder of molecules is more probable. The disorder of molecules is reflected by the entropy of the system. A liquid vapourizes to form a more disordered gas. When a solute is present, there is additional contribution to the entropy of the liquid due to increased randomness. As the entropy of solution is higher than that of pure liquid, there is weaker tendency to form the gas. Thus, a solute (non-volatile) lowers the vapour pressure of a liquid, and hence a higher boiling point of the solution. Similarly, the greater randomness of the solution opposes the tendercy to freeze. In consequence, a lower temperature must be reached for achieving the equilibrium between the solid (frozen solvent) and the solution. The elevation in boiling point (DeltaT_(b)) and depression in freezing point (DeltaT_(f)) of a solution are the colligative properties which depend only on the concentration of particles of the solute and not their identity. For dilute solutions, (DeltaT_(b)) and (DeltaT_(f)) are proportional to the molarity of the solute in the solution. A mixture of two immiscible liquids at a constant pressure of 1.0 atm boils at temperature

A system of greater disorder of molecules is more probable. The disorder of molecules is reflected by the entropy of the system. A liquid vapourizes to form a more disordered gas. When a solute is present, there is additional contribution to the entropy of the liquid due to increased randomness. As the entropy of solution is higher than that of pure liquid, there is weaker tendency to form the gas. Thus, a solute (non-volatile) lowers the vapour pressure of a liquid, and hence a higher boiling point of the solution. Similarly, the greater randomness of the solution opposes the tendercy to freeze. In consequence, a lower temperature must be reached for achieving the equilibrium between the solid (frozen solvent) and the solution. The elevation in boiling point (DeltaT_(b)) and depression in freezing point (DeltaT_(f)) of a solution are the colligative properties which depend only on the concentration of particles of the solute and not their identity. For dilute solutions, (DeltaT_(b)) and (DeltaT_(f)) are proportional to the molarity of the solute in the solution. To aqueous solution of Nal , increasing amounts of solid Hgl_(2) is added. The vapour pressure of the solution

Entropy of the solutions is higher than that of pure liquid . Why?

entropy of a system decreases

Why vapour pressure of a liquid decreases when a non - volatile solute is added to it ?

The solubility of gas in liquid is directly proportional to the pressure over the solutions at a given temperature.

Dissolution of a non-volatile solute into a liquid leads to the-

A solution of two liquids boils at a temperature more than the boiling point of either of them. Hence, the binary solution shows

The absolute entropy of a liquid is less than of the solid.

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