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

1. Benzene + Toluene

2. Acetone + Chloroform

3. Chloroethane + Bromoethane

4. Ethanol + Acetone

The freezing point of depression constant (K_f ) of benzene is 5.12 K kg mol^-1. The freezing point depression for the solution of molality 0.078 m containing a non-electrolyte solute in benzene is-
(rounded off upto two decimal places):
1. 0.80 K
2. 0.40 K
3. 0.60 K
4. 0.20 K

If 8 g of a non-electrolyte solute is dissolved in 114 g of n-octane to reduce its vapor pressure to 80 %, the molar mass (in g mol^–1) of the solute is -

[Molar mass of n-octane is 114 g mol^–1]

Isotonic solutions have the same:

1. Vapour pressure

2. Freezing temperature

3. Osmotic pressure

4. Boiling temperature

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

The following solutions were prepared by dissolving 10 g of glucose (C₆H₁₂O₆)
in 250 ml of water (P₁), 10 g of urea (CH₄N₂O) in 250 ml of water (P₂) and 10 g
of sucrose (C₁₂H₂₂O₁₁) in 250 ml of water (P₃). The decreasing order of osmotic
the pressure of these solutions is:

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 among the following that forms maximum boiling azeotrope is:

1. Heptane + Octane
2. Water + Nitric acid
3. Ethanol + Water
4. Acetone + Carbon disulfide

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