The boiling point of 0.2 mol kg-1 solution of X in water is greater than the equimolal solution of Y in water. The correct statement in this case is:

1. X is undergoing dissociation in water.
2. Molecular mass of X is greater than the molecular mass of Y.
3. Molecular mass of X is less than the molecular mass of Y.
4. Y is undergoing dissociation in water while X undergoes no change.

Subtopic:  Elevation of Boiling Point |
 59%
From NCERT
NEET - 2015
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NEET 2023 - Target Batch - Aryan Raj Singh
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The electrolyte having the same value of Van't Hoff factor (i) as that of Al2(SO4)3  (if all are 100% ionized) is:

1. K2SO4
2. K3[Fe(CN)6]
3. Al(NO3)3
4. K4[Fe(CN)6]

Subtopic:  Introduction & Colligative properties | Van’t Hoff Factor |
 85%
From NCERT
NEET - 2015
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NEET 2023 - Target Batch - Aryan Raj Singh
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The largest freezing point depression among the following O.10 m solutions is shown by:

1. \(\mathrm{KCl}\) 2. \(\mathrm{C_6H_{12}O_6}\)
3. \(\mathrm{Al}_2(\mathrm{SO_4})_3\) 4. \(\mathrm{K_2SO_4}\)
Subtopic:  Relative Lowering of Vapour Pressure |
 80%
From NCERT
AIPMT - 2014
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NEET 2023 - Target Batch - Aryan Raj Singh
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pA and pB are the vapor pressure of pure liquid components, A and B, respectively of an ideal binary solution.
If XA represents the mole fraction of component A, the total pressure of the solution will be:

1. pA + XA (pB-pA)
2. pA + XA(pA-pB)
3. pB + XA(pB-pA)
4. pB + XA(pA-pB)

Subtopic:  Dalton’s Law of Partial Pressure |
 70%
From NCERT
AIPMT - 2012
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The freezing point depression constant for water is 1.86 oC m-1. If 5.00 g Na2SOis dissolved in 45.0 g H2O, the freezing point is changed by -3.82 oC. The van’t Hoff factor for Na2SO4 is:

1. 2.63 2. 3.11
3. 0.381 4. 2.05
Subtopic:  Depression of Freezing Point |
 69%
From NCERT
AIPMT - 2011
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NEET 2023 - Target Batch - Aryan Raj Singh
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The van’t Hoff factor, i, for a compound that undergoes
dissociation and association in a solvent is, respectively:

1. Less than one and less than one.
2. Greater than one and less than one.
3. Greater than one and greater than one.
4. Less than one and greater than one.

Subtopic:  Van’t Hoff Factor |
 85%
From NCERT
AIPMT - 2011
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NEET 2023 - Target Batch - Aryan Raj Singh
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An aqueous solution is 1.00 molal in KI. The vapour pressure of the solution can be increased by:

1. Addition of NaCl

2. Addition of Na2SO4

3. Addition of 1.00  molal Kl

4. Addition of water

Subtopic:  Relative Lowering of Vapour Pressure |
 62%
From NCERT
AIPMT - 2010
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A solution of sucrose (molar mass = 342 g mol-1 ) has been prepared by dissolving 68.5 g of sucrose in 1000 g of water. The freezing point of the solution obtained will be: 
(kf for water = 1.86 K kg mol-1)
1. -0.372 oC
2. -0.520 oC
3. +0.372 oC
4. -0.570 oC

Subtopic:  Depression of Freezing Point |
 81%
From NCERT
AIPMT - 2010
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A 0.0020 m aqueous solution of an ionic compound Co(NH3)5(NO2)Cl freezes at -0.0073 oC. The number of moles of ions that 1 mol of ionic compound produces on being dissolved in water will be:
(Kf = -1.86 oC/m)

1. 2 2. 3
3. 4 4. 1
Subtopic:  Depression of Freezing Point | Van’t Hoff Factor |
 66%
From NCERT
AIPMT - 2009
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NEET 2023 - Target Batch - Aryan Raj Singh
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0.5 molal aqueous solution of a weak acid (HX) is 20 % ionised. The lowering in the freezing point of the solution will be:
[Kf for water = 1.86 K kg mol-1]

1. -1.12 K

2. 0.56 K

3. 1.12 K

4. -0.56 K

Subtopic:  Depression of Freezing Point | Van’t Hoff Factor |
 54%
From NCERT
AIPMT - 2007
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NEET 2023 - Target Batch - Aryan Raj Singh
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