For an aqueous solution, freezing point is -0.186oC. Elevation of the boiling point of the same solution is:
(Given: Kf =1.86 K kg mol-1  and Kb = 0.512 K kg mol-1

1. 0.186

2. 0.0512

3. 1.86

4. 5.12

Subtopic:  Elevation of Boiling Point | Depression of Freezing Point |
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The elevation in boiling point for 1 molal solution of a non-volatile solute A is 3 K. The depression in the freezing point for 2 molal solutions of A in the same solvent is 6 K. The ratio of \(K_b \text { and } K_f \text { i.e., } K_b / K_f~\text{is}~1:X.\) 
The value of \(X\) is:
1. 7 2. 3
3. 1 4. 2
Subtopic:  Elevation of Boiling Point | Depression of Freezing Point |
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JEE
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The depression in freezing point of 0.01 M aqueous solutions of urea, sodium chloride and sodium sulphate is in the ratio of:

1. 1 : 1 : 1

2. 1 : 2 : 3

3. 1 : 2 : 4

4. 2 : 2 : 3

Subtopic:  Depression of Freezing Point |
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What is the molality of the solution of a certain solute in a solvent if there is a freezing point depression of 0.184° and if the freezing point constant is 18.4 K kg 

1. 0.01

2. 1.00

3. 0.001

4. 100

Subtopic:  Depression of Freezing Point |
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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 |
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NEET - 2006

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If the molality of the dilute solution is doubled, then the value of a molal depression constant (Kf) will be:

1. Halved

2. Tripled

3. Unchanged

4. Doubled

Subtopic:  Depression of Freezing Point |
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When 5 g of non-volatile non-electrolyte solute is dissolved in 100 g of a certain solvent, the freezing point of the solvent decreases by 0.25 K. The molar mass of the solute is:
\([\text{K}_f \) of the given solvent \( =1.2 \mathrm{K ~kg}~ \mathrm{mol}^{-1}] \)
1. \(242.8 \mathrm{~g} \mathrm{~mol}^{-1}\) 2. \(238.2 \mathrm{~g} \mathrm{~mol}^{-1}\)
3. \(241.8 \mathrm{~g} \mathrm{~mol}^{-1}\) 4. \(240.0 \mathrm{~g} \mathrm{~mol}^{-1}\)
Subtopic:  Depression of Freezing Point |
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NEET - 2024

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The molecular weight of benzoic acid in benzene, as determined by the depression in the freezing point method, corresponds to:
1. Ionization of benzoic acid
2. Dimerization of benzoic acid
3. Trimerization of benzoic acid
4. Solvation of benzoic acid
Subtopic:  Depression of Freezing Point |
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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 |
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AIPMT - 2010

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The molar freezing point constant for water is 1.86°C/mole. If 342 g of cane sugar (C12H22O11) is dissolved in 1000 g of water, the solution will freeze at 

1. –1.86°C

2. 1.86°C

3. –3.92°C

4. 2.42°C

Subtopic:  Depression of Freezing Point |
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