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प्रश्न
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.
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उत्तर
The melting point of ice is the freezing point of water. We can use the depression in freezing point property in this case.
a. 3rd reading for 0.5 g there has to be an increase in the depression of freezing point and therefore decrease in freezing point so also decrease in melting point when the amount of salt is increased but the trend is not followed in this case.
b. Two sets of reading help to avoid errors in data collection and give more objective data.
c. ΔTf (glucose) = `1 xx "K"_"f" xx (0.6 xx 1000)/(180 xx 10)`
ΔTf (NaCl) = `2 xx "K"_"f" xx (0.6 xx 1000)/(58.5 xx 10)`
3.8 = `2 xx "K"_"f" xx (0.6 xx 1000)/(58.5 xx 10)`
Divide equation 1 by 2
`(Δ"T"_"f" ("glucose"))/3.8 = 58.5/(2 xx 180)`
ΔTf (glucose) = 0.62
Freezing point or Melting point = - 0.62°C
OR
Depression in freezing point is directly proportional to molality (mass of solute when the amount of solvent remains same)
0.3 g depression is 1.9°C
0.6 g depression is 3.8°C
1.2 g depression will be 3.8 × 2 = 7.6°C
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संबंधित प्रश्न
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
A 4% solution(w/w) of sucrose (M = 342 g mol–1) in water has a freezing point of 271.15 K. Calculate the freezing point of 5% glucose (M = 180 g mol–1) in water.
(Given: Freezing point of pure water = 273.15 K)
Cryoscopic constant of a liquid is ____________.
Which of the following 0.10 m aqueous solutions will have the lowest freezing point?
Which observation(s) reflect(s) colligative properties?
(i) A 0.5 m NaBr solution has a higher vapour pressure than a 0.5 m BaCl2 solution at the same temperature.
(ii) Pure water freezes at a higher temperature than pure methanol.
(iii) A 0.1 m NaOH solution freezes at a lower temperature than pure water.
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:
1.2 mL of acetic acid is dissolved in water to make 2.0 L of solution. The depression in freezing point observed for this strength of acid is 0.0198° C. The percentage of dissociation of the acid is ______. [Nearest integer]
[Given: Density of acetic acid is 1.02 g mL–1, Molar mass of acetic acid is 60 g/mol.]
Kf(H2O) = 1.85 K kg mol–1
40 g of glucose (Molar mass = 180) is mixed with 200 mL of water. The freezing point of solution is ______ K. (Nearest integer)
[Given : Kf = 1.86 K kg mol-1; Density of water = 1.00 g cm-3; Freezing point of water = 273.15 K]
Ibrahim collected 10 mL each of fresh water and ocean water. He observed that one sample labeled “P” froze at 0° C while the other “Q” at -1.3° C. Ibrahim forgot which of the two, “P” or “Q” was ocean water. Help him identify which container contains ocean water, giving rationalization for your answer.
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.
