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Question
Calculate the freezing point of the solution when 31 g of ethylene glycol (C2H6O2) is dissolved in 500 g of water.
(Kf for water = 1.86 K kg mol–1)
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Solution
The freezing point of pure water is 273.15 K. On dissolving ethylene glycol, freezing point, being a colligative property, will be lowered.
`DeltaT_f=K_"f"(w_sxx1000)/(M_sxxW`
We are given that
Kf for water=1.86 K kg mol−1
Mass of solute, ws=31 g
Molar mass of solute, Ms = 12 × 2 + 1 × 6 + 16 × 2
= 62 g mol−1
Mass of water, W=500 g
`thereforeDeltaT_f=1.86xx(31xx1000)/(62xx500)`
= 1.86 K
Hence, the freezing point of the solution,
Tf = 273.15 K−1.86 K
= 271.29 K
RELATED QUESTIONS
0.01 M solution of KCl and BaCl2 are prepared in water. The freezing point of KCl is found to be – 2°C. What is the freezing point of BaCl2 to be completely ionised?
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.
Which has the highest 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.
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
1000 g of 1 m sucrose solution in water is cooled to −3.534°C. What weight of ice would be separated out at this temperature 1 is ______ gm. Kf(H2O) = 1.86 K mol−1 Kg)
When 25.6 g of sulphur was dissolved in 1000 g of benzene, the freezing point lowered by 0.512 K. Calculate the formula of sulphur (Sr).
(Kf for benzene = 5.12 K kg mol−1, Atomic mass of sulphur = 32 g mol−1)
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.
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.
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.
