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प्रश्न
State Raoult’s law for the solution containing volatile components
उत्तर
Raoult’s law: In a solution, the vapour pressure of a component at a given temperature is equal to the mole fraction of that component in the solution multiplied by the vapour pressure of that component in the pure state.
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संबंधित प्रश्न
Calculate the mass of a non-volatile solute (molar mass 40 g mol−1) which should be dissolved in 114 g octane to reduce its vapour pressure to 80%.
A solution containing 30 g of non-volatile solute exactly in 90 g of water has a vapour pressure of 2.8 kPa at 298 K. Further, 18 g of water is then added to the solution and the new vapour pressure becomes 2.9 kPa at 298 K. Calculate:
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- vapour pressure of water at 298 K.
100 g of liquid A (molar mass 140 g mol−1) was dissolved in 1000 g of liquid B (molar mass 180 g mol−1). The vapour pressure of pure liquid B was found to be 500 torr. Calculate the vapour pressure of pure liquid A and its vapour pressure in the solution if the total vapour pressure of the solution is 475 Torr.
What type of azeotrope is formed by positive deviation from Raoult's law ? Give an example.
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Match the following:
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| (ii) | Nicol prism | (b) | Osmotic pressure |
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| (iv) | Starch | (d) | Polarimeter |
| (v) | Acetaldehyde | (e) | Arrhenius equation |
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What will be the vapour pressure of a solution containing 5 moles of sucrose (C12H22O11) in 1 kg of water, if the vapour pressure of pure water is 4.57 mm of Hg? [C = 12, H = 1, O = 16]
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On the basis of information given below mark the correct option.
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