1 Inversely proportional to the mole fraction of the solvent.
2 Directly proportional to the mole fraction of the solvent.
3 Directly proportional to the mole fraction of the solute.
4 Inversely proportional to the mole fraction of the solute.
Explanation:
For solutions containing non-volatile solutes, the Raoult’s law may be stated as at a given temperature, the vapour pressure of a solution containing non-volatile solute is directly proportional to the mole fraction of the solvent.
CHXII02:SOLUTIONS
319429
Assertion : Both vapour pressure and boiling point depends on surface area of the liquid. Reason : Higher the surface area, lower be the boiling point whereas higher will be the vapour pressure.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Both Assertion and Reason are incorrect
Explanation:
Both vapour pressure and boiling point are independent of surface area. They have constant values at a given temperature. Hence, Assertion and Reason both are false.
AIIMS - 2017
CHXII02:SOLUTIONS
319430
The vapour pressure of water depends upon
1 surface area of container
2 volume of container
3 temperature
4 All of these
Explanation:
Lowering of vapour pressure depends on number of particles of solute and temperature.
CHXII02:SOLUTIONS
319431
Out of the compounds given below, the vapour pressure of (B) at a particular temperature is:
1 Higher than that of \(\mathrm{A}\)
2 Lower than that of \(\mathrm{B}\)
3 Higher or lower than \({\text{(A)}}\), depending on the size of the vessel
4 Same as that of \({\text{(A)}}\)
Explanation:
In (A) Para-nitrophenol intermolecular \(\mathrm{H}\)-bonding exists, while in \((\mathrm{B})\) orthonitrophenol, intramolecular H-bonding exists. The boiling point of \((\mathrm{B})\) is lower than (A) and, thus, (B) is more volatile, i.e., \((\mathrm{B})\) has higher vapour pressure as compared to \((\mathrm{A})\).
CHXII02:SOLUTIONS
319432
A substance will be deliquescent if its vapour pressure is:
1 equal to that of water vapour in the air
2 lesser than that of water vapour in the air
3 equal to the atmospheric pressure
4 greater than that of water vapour in the air
Explanation:
Higher vapour pressure of \({{\rm{H}}_{\rm{2}}}{\rm{O}}\) in atmosphere will derive \({{\rm{H}}_{\rm{2}}}{\rm{O}}\) vapours to solute particles.
1 Inversely proportional to the mole fraction of the solvent.
2 Directly proportional to the mole fraction of the solvent.
3 Directly proportional to the mole fraction of the solute.
4 Inversely proportional to the mole fraction of the solute.
Explanation:
For solutions containing non-volatile solutes, the Raoult’s law may be stated as at a given temperature, the vapour pressure of a solution containing non-volatile solute is directly proportional to the mole fraction of the solvent.
CHXII02:SOLUTIONS
319429
Assertion : Both vapour pressure and boiling point depends on surface area of the liquid. Reason : Higher the surface area, lower be the boiling point whereas higher will be the vapour pressure.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Both Assertion and Reason are incorrect
Explanation:
Both vapour pressure and boiling point are independent of surface area. They have constant values at a given temperature. Hence, Assertion and Reason both are false.
AIIMS - 2017
CHXII02:SOLUTIONS
319430
The vapour pressure of water depends upon
1 surface area of container
2 volume of container
3 temperature
4 All of these
Explanation:
Lowering of vapour pressure depends on number of particles of solute and temperature.
CHXII02:SOLUTIONS
319431
Out of the compounds given below, the vapour pressure of (B) at a particular temperature is:
1 Higher than that of \(\mathrm{A}\)
2 Lower than that of \(\mathrm{B}\)
3 Higher or lower than \({\text{(A)}}\), depending on the size of the vessel
4 Same as that of \({\text{(A)}}\)
Explanation:
In (A) Para-nitrophenol intermolecular \(\mathrm{H}\)-bonding exists, while in \((\mathrm{B})\) orthonitrophenol, intramolecular H-bonding exists. The boiling point of \((\mathrm{B})\) is lower than (A) and, thus, (B) is more volatile, i.e., \((\mathrm{B})\) has higher vapour pressure as compared to \((\mathrm{A})\).
CHXII02:SOLUTIONS
319432
A substance will be deliquescent if its vapour pressure is:
1 equal to that of water vapour in the air
2 lesser than that of water vapour in the air
3 equal to the atmospheric pressure
4 greater than that of water vapour in the air
Explanation:
Higher vapour pressure of \({{\rm{H}}_{\rm{2}}}{\rm{O}}\) in atmosphere will derive \({{\rm{H}}_{\rm{2}}}{\rm{O}}\) vapours to solute particles.
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
CHXII02:SOLUTIONS
319428
Vapour pressure of a solution is
1 Inversely proportional to the mole fraction of the solvent.
2 Directly proportional to the mole fraction of the solvent.
3 Directly proportional to the mole fraction of the solute.
4 Inversely proportional to the mole fraction of the solute.
Explanation:
For solutions containing non-volatile solutes, the Raoult’s law may be stated as at a given temperature, the vapour pressure of a solution containing non-volatile solute is directly proportional to the mole fraction of the solvent.
CHXII02:SOLUTIONS
319429
Assertion : Both vapour pressure and boiling point depends on surface area of the liquid. Reason : Higher the surface area, lower be the boiling point whereas higher will be the vapour pressure.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Both Assertion and Reason are incorrect
Explanation:
Both vapour pressure and boiling point are independent of surface area. They have constant values at a given temperature. Hence, Assertion and Reason both are false.
AIIMS - 2017
CHXII02:SOLUTIONS
319430
The vapour pressure of water depends upon
1 surface area of container
2 volume of container
3 temperature
4 All of these
Explanation:
Lowering of vapour pressure depends on number of particles of solute and temperature.
CHXII02:SOLUTIONS
319431
Out of the compounds given below, the vapour pressure of (B) at a particular temperature is:
1 Higher than that of \(\mathrm{A}\)
2 Lower than that of \(\mathrm{B}\)
3 Higher or lower than \({\text{(A)}}\), depending on the size of the vessel
4 Same as that of \({\text{(A)}}\)
Explanation:
In (A) Para-nitrophenol intermolecular \(\mathrm{H}\)-bonding exists, while in \((\mathrm{B})\) orthonitrophenol, intramolecular H-bonding exists. The boiling point of \((\mathrm{B})\) is lower than (A) and, thus, (B) is more volatile, i.e., \((\mathrm{B})\) has higher vapour pressure as compared to \((\mathrm{A})\).
CHXII02:SOLUTIONS
319432
A substance will be deliquescent if its vapour pressure is:
1 equal to that of water vapour in the air
2 lesser than that of water vapour in the air
3 equal to the atmospheric pressure
4 greater than that of water vapour in the air
Explanation:
Higher vapour pressure of \({{\rm{H}}_{\rm{2}}}{\rm{O}}\) in atmosphere will derive \({{\rm{H}}_{\rm{2}}}{\rm{O}}\) vapours to solute particles.
1 Inversely proportional to the mole fraction of the solvent.
2 Directly proportional to the mole fraction of the solvent.
3 Directly proportional to the mole fraction of the solute.
4 Inversely proportional to the mole fraction of the solute.
Explanation:
For solutions containing non-volatile solutes, the Raoult’s law may be stated as at a given temperature, the vapour pressure of a solution containing non-volatile solute is directly proportional to the mole fraction of the solvent.
CHXII02:SOLUTIONS
319429
Assertion : Both vapour pressure and boiling point depends on surface area of the liquid. Reason : Higher the surface area, lower be the boiling point whereas higher will be the vapour pressure.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Both Assertion and Reason are incorrect
Explanation:
Both vapour pressure and boiling point are independent of surface area. They have constant values at a given temperature. Hence, Assertion and Reason both are false.
AIIMS - 2017
CHXII02:SOLUTIONS
319430
The vapour pressure of water depends upon
1 surface area of container
2 volume of container
3 temperature
4 All of these
Explanation:
Lowering of vapour pressure depends on number of particles of solute and temperature.
CHXII02:SOLUTIONS
319431
Out of the compounds given below, the vapour pressure of (B) at a particular temperature is:
1 Higher than that of \(\mathrm{A}\)
2 Lower than that of \(\mathrm{B}\)
3 Higher or lower than \({\text{(A)}}\), depending on the size of the vessel
4 Same as that of \({\text{(A)}}\)
Explanation:
In (A) Para-nitrophenol intermolecular \(\mathrm{H}\)-bonding exists, while in \((\mathrm{B})\) orthonitrophenol, intramolecular H-bonding exists. The boiling point of \((\mathrm{B})\) is lower than (A) and, thus, (B) is more volatile, i.e., \((\mathrm{B})\) has higher vapour pressure as compared to \((\mathrm{A})\).
CHXII02:SOLUTIONS
319432
A substance will be deliquescent if its vapour pressure is:
1 equal to that of water vapour in the air
2 lesser than that of water vapour in the air
3 equal to the atmospheric pressure
4 greater than that of water vapour in the air
Explanation:
Higher vapour pressure of \({{\rm{H}}_{\rm{2}}}{\rm{O}}\) in atmosphere will derive \({{\rm{H}}_{\rm{2}}}{\rm{O}}\) vapours to solute particles.
1 Inversely proportional to the mole fraction of the solvent.
2 Directly proportional to the mole fraction of the solvent.
3 Directly proportional to the mole fraction of the solute.
4 Inversely proportional to the mole fraction of the solute.
Explanation:
For solutions containing non-volatile solutes, the Raoult’s law may be stated as at a given temperature, the vapour pressure of a solution containing non-volatile solute is directly proportional to the mole fraction of the solvent.
CHXII02:SOLUTIONS
319429
Assertion : Both vapour pressure and boiling point depends on surface area of the liquid. Reason : Higher the surface area, lower be the boiling point whereas higher will be the vapour pressure.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Both Assertion and Reason are incorrect
Explanation:
Both vapour pressure and boiling point are independent of surface area. They have constant values at a given temperature. Hence, Assertion and Reason both are false.
AIIMS - 2017
CHXII02:SOLUTIONS
319430
The vapour pressure of water depends upon
1 surface area of container
2 volume of container
3 temperature
4 All of these
Explanation:
Lowering of vapour pressure depends on number of particles of solute and temperature.
CHXII02:SOLUTIONS
319431
Out of the compounds given below, the vapour pressure of (B) at a particular temperature is:
1 Higher than that of \(\mathrm{A}\)
2 Lower than that of \(\mathrm{B}\)
3 Higher or lower than \({\text{(A)}}\), depending on the size of the vessel
4 Same as that of \({\text{(A)}}\)
Explanation:
In (A) Para-nitrophenol intermolecular \(\mathrm{H}\)-bonding exists, while in \((\mathrm{B})\) orthonitrophenol, intramolecular H-bonding exists. The boiling point of \((\mathrm{B})\) is lower than (A) and, thus, (B) is more volatile, i.e., \((\mathrm{B})\) has higher vapour pressure as compared to \((\mathrm{A})\).
CHXII02:SOLUTIONS
319432
A substance will be deliquescent if its vapour pressure is:
1 equal to that of water vapour in the air
2 lesser than that of water vapour in the air
3 equal to the atmospheric pressure
4 greater than that of water vapour in the air
Explanation:
Higher vapour pressure of \({{\rm{H}}_{\rm{2}}}{\rm{O}}\) in atmosphere will derive \({{\rm{H}}_{\rm{2}}}{\rm{O}}\) vapours to solute particles.