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प्रश्न
Out of the following 1.0 M aqueous solution, which one will show the largest freezing point depression?
विकल्प
NaCl
Na2SO4
C6H12O6
Al2(SO4)3
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उत्तर
Al2(SO4)3
Explanation:
Freezing point depression is given by the formula;
ΔTf = i × Kb × m
Where, ΔTf = Freezing point depression
Tf = Freezing point of the solution (°C)
i = ionization or van't Hoff factor
Kf = cryoscopic constant (°C/m)
m = molality (mol/kg)
Let's check for each option one by one.
(1) \[\ce{KCl_{(aq)} <=> K^+_{ (aq)} + Cl^-_{ (aq)}}\]
total ions = 2 thus, i = 2
(2) Glucose is non-electrolyte, so it does not dissociate into ions.
\[\ce{C6H12O6 <=> no ions}\] [i = 0]
(3) \[\ce{Al2(SO4)3_{(aq)} <=> 2Al^{3+} + 3SO^{2-}_4}\]
total ions = 5 thus, i = 5
(4) \[\ce{K2SO4_{ (aq)} <=> 2K^+ + SO^{2-}_4}\]
total ions = 3 thus, i = 3
Hence, Al2(SO4)3 will exhibit the largest freezing point depression due to the highest value of i.
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संबंधित प्रश्न
Define Cryoscopic constant.
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
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:
A 5% solution (by mass) of cane sugar in water has a freezing point of 271 K and a freezing point of pure water is 273.15 K. The freezing point of a 5% solution (by mass) of glucose in water is:
The freezing point of equimolal aqueous solution will be highest for ____________.
Given below are two statements labelled as Assertion (A) and Reason (R).
Assertion (A): Cryoscopic constant depends on nature of solvent.
Reason (R): Cryoscopic constant is a universal constant.
Select the most appropriate answer from the options given below:
Which of the following statements is false?
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.
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 |
Assuming the melting point of pure water as 0°C, answer the following questions:
- One temperature in the second set of results does not fit the pattern. Which temperature is that? Justify your answer.
- 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.
The depression in the 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.
