The correct option for the value of vapor pressure of a solution at 45 $°$C with benzene to octane in a molar ratio 3:2 is:

[At 45 $°$C vapor pressure of benzene is 280 mm Hg and that of octane is 420 mm Hg. Assume Ideal gas]
1. 336 mm of Hg
2. 350 mm of Hg
3. 160 mm of Hg
4. 168 mm of Hg

The mixture that shows positive deviation from Raoult's law is-

1. Benzene + Toluene

2. Acetone + Chloroform

3. Chloroethane + Bromoethane

4. Ethanol + Acetone

For an ideal solution, the correct option is -

1. ${∆}_{\mathrm{mix}}$ $\mathrm{G}=0$ at constant T and P

2. ${∆}_{\mathrm{mix}}$ $\mathrm{S}=0$ at constant T and P

3. ${∆}_{\mathrm{mix}}$ $\mathrm{V}\ne 0$ at constant T and P

4. ${∆}_{\mathrm{mix}}$ $\mathrm{H}=0$ at constant T and P

The correct statement regarding a solution of two components A and B exhibiting positive deviation from ideal behavior is :

Vapour pressure of chloroform $\left(CHC{l}_3\right)$ and dichloromethane $\left(C{H}_{2}C{l}_2\right)$ at $25°C$ are 200 mmHg and 41.5 mmHg respectively. Vapour pressure of the solution was obtained by mixing 25.5 g of $\left(CHC{l}_3\right)$ and 40 g of $\left(C{H}_{2}C{l}_2\right)$ at the same temperature will be: (Molecular mass of $\left(CHC{l}_3\right)$ = 119.5 u and molecular mass of $\left(C{H}_{2}C{l}_2\right)$ = 85 u)

A solution of acetone in ethanol:

1. Shows a negative deviation from Raoult's law

2. Shows a positive deviation from Raoult's law

3. Behaves like a near-ideal solution

4. Obeys Raoult's law

The vapour pressure of two liquids 'P' and 'Q' are 80 and 60 torr, respectively. The total vapour pressure of the solution obtained by mixing 3 moles of P and 2 moles of Q would be:

1. 68 torr

2. 140 torr

3. 72 torr

4. 20 torr

A solution has a 1:4 mole ratio of pentane to hexane.
The vapor pressures of the pure hydrocarbons at 20 ºC are
440 mm Hg for pentane and 120 mm Hg for hexane.
The mole fraction of pentane in the vapor phase would be:

The ideal solution indicates: