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

1. Benzene + Toluene

2. Acetone + Chloroform

3. Chloroethane + Bromoethane

4. Ethanol + Acetone

Subtopic:  Raoult's Law |
67%
From NCERT
NEET - 2020
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The correct option for the value of vapour pressure of a solution at 45 $°$C with benzene to octane in a molar ratio 3:2 is:

[At 45 $°$C vapour 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

Subtopic:  Raoult's Law |
73%
From NCERT
NEET - 2021
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Vapour pressure of chloroform (CHCl3) and dichloromethane (CH2Cl2at 298 K are 200 mm Hg and 415 mm Hg respectively. The vapour pressure of the solution prepared by mixing 25.5 g of CHCl3 and 40 g of CH2Cl2 at 298 K is 347.9 mm Hg.

The mole fractions of each component, that is, CHCl
3 and CH2Cl2 in vapour phase are respectively:

0.82; 0.18
2. 0.16; 0.84
3. 0.18; 0.82
4. 0.84; 0.16

Subtopic:  Raoult's Law |
From NCERT
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On mixing, heptane, and octane form an ideal solution at 373 K, the vapor pressures of the two liquid components (Heptane and octane) are 105 kPa and 45 kPa respectively. Vapour pressure of the solution obtained by mixing 25.0 g of heptane and 35 g of octane will be:
(molar mass of heptane = 100 g mol−1 and of octane =114 g 1 mol−1)

1. 144.5 kPa

2. 72.0 kPa

3. 36.1 kPa

4. 96.2 kPa

Subtopic:  Raoult's Law |
76%
From NCERT
 1 Raoult's law states that the vapour pressure of a component over a solution is proportional to its mole fraction. 2 The osmotic pressure ($$\pi$$) of a solution is given by the equation $$\pi$$=MRT, where M is the molarity of the solution. 3 The correct order of osmotic pressure for 0.10 M aqueous solution of each compound is BaCl2 > KCl > CH3COOH > sucrose. 4 Two sucrose solutions of the same molarity prepared in different solvents will have the same depression in the freezing point.