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प्रश्न
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)
उत्तर
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
संबंधित प्रश्न
Calculate the amount of CaCl2 (molar mass = 111 g mol−1) which must be added to 500 g of water to lower its freezing point by 2 K, assuming CaCl2 is completely dissociated. (Kf for water = 1.86 K kg mol−1)
Cryoscopic constant of a liquid 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 ____________.
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.
Which of the following statement is false?
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.
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]
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.
