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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)
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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 freezing point of solution when 1.9 g of MgCl2 (M = 95 g mol−1) was dissolved in 50 g of water, assuming MgCl2 undergoes complete ionization.
(Kf for water = 1.86 K kg mol−1)
Calculate the mass of ascorbic acid (Vitamin C, C6H8O6) to be dissolved in 75 g of acetic acid to lower its melting point by 1.5°C. Kf = 3.9 K kg mol−1.
Calculate the depression in the freezing point of water when 10 g of CH3CH2CHClCOOH is added to 250 g of water. Ka = 1.4 × 10−3, Kf = 1.86 K kg mol−1.
Calculate the freezing point of a solution containing 60 g of glucose (Molar mass = 180 g mol–1) in 250 g of water. (Kf of water = 1.86 K kg mol–1)
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)
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)
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 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.
