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%
Level 2: 60%+
NEET - 2009
Hints

Kohlrausch's law states that at

(1) finite dilution, each ion makes definite contribution to equivalent conductance of an electrolyte, whatever be the nature of the other ion of the electrolyte.

(2) infinite dilution, each ion makes definite contribution to equivalent conductance of an electrolyte depending on the nature of the other ion of th electrolyte.

(3) infinite dilution, each ion makes definite contribution to conductance of an electrolyte whatever be the nature of the other ion of the electrolyte.

(4) infinite dilution, each ion makes definite contribution to equivalent conductance of an electrolyte, whatever be the nature of the other ion of the electrolyte.

Subtopic:   Kohlrausch Law & Cell Constant |
 50%
Level 3: 35%-60%
NEET - 2008
Hints

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%
Level 3: 35%-60%
NEET - 2007
Hints

advertisementadvertisement

 A solution containing 10 g per dm3 of urea (molecular mass = 60 g mol-1) is isotonic with a 5% solution of a non-volatile solute. The molecular mass of this non-volatile solute is:

(1) 250 g mol-1               

(2) 300 g mol-1

(3) 350 g mol-1               

(4) 200 g mol-1

Subtopic:  Concentration Terms & Henry's Law |
 82%
Level 1: 80%+
NEET - 2006
Hints

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 |
 80%
Level 1: 80%+
NEET - 2006
Hints

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 |
 75%
Level 2: 60%+
NEET - 2006
Hints

advertisementadvertisement

premium feature crown icon
Unlock IMPORTANT QUESTION
This question was bookmarked by 5 NEET 2025 toppers during their NEETprep journey. Get Target Batch to see this question.
✨ Perfect for quick revision & accuracy boost
Buy Target Batch
Access all premium questions instantly

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 |
Level 3: 35%-60%
Hints

premium feature crown icon
Unlock IMPORTANT QUESTION
This question was bookmarked by 5 NEET 2025 toppers during their NEETprep journey. Get Target Batch to see this question.
✨ Perfect for quick revision & accuracy boost
Buy Target Batch
Access all premium questions instantly

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%
Level 3: 35%-60%
Hints

premium feature crown icon
Unlock IMPORTANT QUESTION
This question was bookmarked by 5 NEET 2025 toppers during their NEETprep journey. Get Target Batch to see this question.
✨ Perfect for quick revision & accuracy boost
Buy Target Batch
Access all premium questions instantly

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 |
 51%
Level 3: 35%-60%
Hints

advertisementadvertisement

premium feature crown icon
Unlock IMPORTANT QUESTION
This question was bookmarked by 5 NEET 2025 toppers during their NEETprep journey. Get Target Batch to see this question.
✨ Perfect for quick revision & accuracy boost
Buy Target Batch
Access all premium questions instantly

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 |
Level 3: 35%-60%
Hints