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प्रश्न
Define Cryoscopic constant.
उत्तर
Cryoscopic constant or the Molal depression constant is defined as the depression in freezing point when one mole of non-volatile solute is dissolved in one kilogram of solvent. Its unit is K.Kg.mol-1.
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संबंधित प्रश्न
1.0 x10-3Kg of urea when dissolved in 0.0985 Kg of a solvent, decreases freezing point of the solvent by 0.211 k. 1.6x10 Kg of another non-electrolyte solute when dissolved in 0.086 Kg of the same solvent depresses the freezing point by 0.34 K. Calculate the molar mass of the another solute. (Given molar mass of urea = 60)
Which of the following solutions shows maximum depression in freezing point?
(A) 0.5 M Li2SQ4
(B) 1 M NaCl
(C) 0.5 M A12(SO4)3
(D) 0.5 MBaC12
When 2.56 g of sulphur was dissolved in 100 g of CS2, the freezing point lowered by 0.383 K. Calculate the formula of sulphur (Sx).
(Kf for CS2 = 3.83 K kg mol−1, Atomic mass of sulphur = 32 g mol−1]
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Define Freezing point.
Calculate the freezing point of a solution containing 60 g of glucose (Molar mass = 180 g mol–1) in 250 g of water. (Kf of water = 1.86 K kg mol–1)
What is the freezing point of a liquid? The freezing point of pure benzene is 278.4 K. Calculate the freezing point of the solution when 2.0 g of a solute having molecular weight 100 g mol-1
is added to 100 g of benzene.
( Kf of benzene = 5.12 K kg mol-1 .)
Pure benzene freezes at 5.45°C. A 0.374 m solution of tetrachloroethane in benzene freezes at 3.55°C. The Kf for benzene is:
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In comparison to a 0.01 m solution of glucose, the depression in freezing point of a 0.01 m MgCl2 solution is ______.
Which of the following statement is false?
Which has the highest freezing point?
How does sprinkling of salt help in clearing the snow covered roads in hilly areas? Explain the phenomenon involved in the process.
Assertion: When NaCl is added to water a depression in freezing point is observed.
Reason: The lowering of vapour pressure of a solution causes depression in the freezing point.
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Read the passage carefully and answer the questions that follow:
Henna is investigating the melting point of different salt solutions. She makes a salt solution using 10 mL of water with a known mass of NaCl salt. She puts the salt solution into a freezer and leaves it to freeze. She takes the frozen salt solution out of the freezer and measures the temperature when the frozen salt solution melts. She repeats each experiment. |
S.No | Mass of the salt used in g |
Melting point in °C | |
Readings Set 1 | Reading Set 2 | ||
1 | 0.3 | -1.9 | -1.9 |
2 | 0.4 | -2.5 | -2.6 |
3 | 0.5 | -3.0 | -5.5 |
4 | 0.6 | -3.8 | -3.8 |
5 | 0.8 | -5.1 | -5.0 |
6 | 1.0 | -6.4 | -6.3 |
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- Why did Henna collect two sets of results?
- In place of NaCl, if Henna had used glucose, what would have been the melting point of the solution with 0.6 g glucose in it?
OR
What is the predicted melting point if 1.2 g of salt is added to 10 mL of water? Justify your answer.
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The depression in freezing point of water observed for the same amount of acetic acid, trichloroacetic acid and trifluoroacetic acid increases in the order given above. Explain briefly.
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