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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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(iii) A 0.1 m NaOH solution freezes at a lower temperature than pure water.
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
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- 0.10 M Na2 SO4
- 0.10 M KCl
- 0.10 M Li3 PO4
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