An aqueous solution is 1.00 molal in KI. Which change will cause the vapour pressure of

the solution to increase ?

1. Addition of NaCl

2. Addition of Na2SO4

3. Addition of 1.00 molal KI

4. Addition of water

Subtopic:  Relative Lowering of Vapour Pressure |
 55%
From NCERT
NEET - 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 °C

2. 0.372 °C

3. 0.572 °C

4. -0.572 °C

Subtopic:  Depression of Freezing Point |
 71%
From NCERT
NEET - 2010
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A 0.0020 m aqueous solution of an ionic compound Co(NH3)5(NO2)Cl freezes at -0.00732°C. Number of moles of ions which 1 mol of ionic compound produces on being dissolved in water will be (kf=-1.86°C/ m)

1. 2

2. 3

3. 4

4. 1

 

Subtopic:  Depression of Freezing Point |
 66%
From NCERT
NEET - 2009
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0.5 molal aqueous solution of a weak acid (HX) is 20% ionised. If Kf for water is 1.86 K kg mol-1, the lowering in freezing point of the solution is:

(1) -1.12 K

(2) 0.56 K

(3) 1.12 K

(4) - 0.56 K

Subtopic:  Depression of Freezing Point |
 50%
From NCERT
NEET - 2007
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1.00 g of a non-electrolyte solute (molar mass 250g/mol) was dissolved in 51.2 g of benzene. If the freezing point depression constant, Kf of benzene is 5.12 K kg mol-1, the freezing point of benzene will be lowered by:-

(1) 0.4 K

(2) 0.3 K

(3) 0.5 K

(4) 0.2 K

Subtopic:  Depression of Freezing Point |
 79%
From NCERT
NEET - 2006
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A solution of acetone in ethanol:

1. Shows a negative deviation from Raoult's law.

2. Shows a positive deviation from Raoult's law.

3. Behaves like a near ideal solution.

4. Obeys Raoult's law.

Subtopic:  Raoult's Law |
 73%
From NCERT
NEET - 2006
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Three solutions are prepared by adding 'w' gm of 'A' into 1kg of water, 'w' gm of 'B' into another 1 kg of water and 'w' gm of 'C' in another 1 kg of water (A, B, C are non electrolytic). Dry air is passed from these solutions in sequence (A → B → C). The loss in weight of solution A was found to be 2gm while solution B gained 0.5 gm and solution C lost 1 gm. Then the relation between molar masses of A, B and C is :

(1) MA: MB : Mc = 4 : 3 : 5

(2) MA : MB: Mc = 14:13:15

(3) Mc > MA > MB

(4) MB > MA> Mc

Subtopic:  Relative Lowering of Vapour Pressure |
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How many mili moles of sucrose should be dissolved in 500 gms of water so as to get a solution which has a difference of 103.57°C between boiling point and freezing point.

(K1 = 1.86 K Kg mol−1, Kb = 0.52 K Kg mo1−1)

(1) 500 mmoles

(2) 900 mmoles

(3) 750 mmoles

(4) 650 mmoles

Subtopic:  Elevation of Boiling Point |
 58%
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The molar heat of vapourization of toluene is AHv. If its vapour pressure at 315 K is 60 torr & that at 356K is300 torr then AHv = ? (log 2 = 0.3)

(1) 37.5 kJ/mole

(2) 3.75 kJ/mole

(3) 37.5 J/mol

(4) 3.75 J/mole

Subtopic:  Relative Lowering of Vapour Pressure |
 50%
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Relative decrease in vapour pressure of an aqueous solution containing 2 moles [Cu(NH3)3Cl) Cl in 3 moles H2O is 0.50. On reaction with AgNO3, this solution will form (assuming no change in degree of ionisation of substance on adding AgNO3)

(1) 1 mol AgCl

(2) 0.25 mol AgCl

(3) 0.5 mol AgCl

(4) 0.40 mol AgCl

Subtopic:  Relative Lowering of Vapour Pressure |
From NCERT
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