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प्रश्न
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)
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उत्तर
Given,
W2 = 60 g
M2 = 180 g mol-1
W1 = 250 g
Kf = 1.86 K kg mol–1
ΔTf = kfm
`T_f^0 - T_f = (K_f xx w_2 xx 1000)/M_2xxw_1`
`2.73 - T_f = (1.86 xx 60 xx 1000)/(180xx250)`
`2.73 - T_f = (1.86 xx 60 xx 1000)/(180 xx 250)`
273.15 - Tf = 2.48
Tf = 273.15 - 2.48 = 270.67 K
Hence, the freezing point of water is 270.67 K or -2.48°C
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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)
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)
Give reasons for the following:
Measurement of osmotic pressure method is preferred for the determination of molar masses of macromolecules such as proteins and polymers.
A solution containing 1.8 g of a compound (empirical formula CH2O) in 40 g of water is observed to freeze at –0.465° C. The molecular formula of the compound is (Kf of water = 1.86 kg K mol–1):
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.
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
What is the predicted melting point if 1.2 g of salt is added to 10 mL of water? Justify your answer.
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)
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]
Out of the following 1.0 M aqueous solution, which one will show the largest freezing point depression?
