Advertisements
Advertisements
Question
What is the freezing point of a liquid? The freezing point of pure benzene is 278.4 K. Calculate the freezing point of the solution when 2.0 g of a solute having molecular weight 100 g mol-1
is added to 100 g of benzene.
( Kf of benzene = 5.12 K kg mol-1 .)
Advertisements
Solution
a) Freezing point: The freezing point of a liquid may be defined as the temperature at which the vapour pressure of solid is equal to the vapour pressure of liquid. A liquid freezes at a temperature at which the liquid and its solid coexist in equilibrium.
b) Solution:
Given:
Freezing point of pure solvent ( T° ) = 278.4 K
Mass of solute ( W2 ) = 2 g = 2 x 10-3 kg
Molar mass of solute ( M2 ) = 100 g mol-1 = 100 x 10-3 kg mol-1
Mass of solvent ( W1) = 100 g = 100 x 10-3 kg
Molal depression constant ( Kf )= 5.12 K kg mol-1.
To find: Freezing point of solution (T)
Formulae:
1. ΔTf = T° - T
2. ΔTf = `( "K"_"f" "W"_2 )/( "M"_2 "W"_1 )`
Calculation:
From formula (2),
ΔTf = `( "K"_"f" "W"_2 )/( "M"_2 "W"_1 )`
ΔTf = `( 5.12 xx 2 xx 10^-3 )/( 100 xx 10^-3 xx 100 xx 10^-3 )`
= `( 5.12 xx 2 )/10`
= 1.024 K
From formula (1),
ΔTf = T° - T
∴ T = T° - ΔTf
∴ T = 278.4 - 1.024 = 277.376 K
∴ Freezing point of the given benzene solution is 277.376 K.
APPEARS IN
RELATED QUESTIONS
Write the formula to determine the molar mass of a solute using freezing point depresssion method.
Define Cryoscopic constant.
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
When 2.56 g of sulphur was dissolved in 100 g of CS2, the freezing point lowered by 0.383 K. Calculate the formula of sulphur (Sx).
(Kf for CS2 = 3.83 K kg mol−1, Atomic mass of sulphur = 32 g 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.
A 5% solution (by mass) of cane sugar in water has freezing point of 271 K. Calculate the freezing point of 5% glucose in water if freezing point of pure water is 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.
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)
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)
Pure benzene freezes at 5.45°C. A 0.374 m solution of tetrachloroethane in benzene freezes at 3.55°C. The Kf for benzene 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:
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?
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):
If molality of dilute solution is doubled, the value of molal depression constant (Kf) will be ______.
Which has the highest freezing point?
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:
How does sprinkling of salt help in clearing the snow covered roads in hilly areas? Explain the phenomenon involved in the process.
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.
Latent heat of water (ice) is 1436.3 cal per mol. What will be molal freezing point depression constant of water (R = 2 cal/degree/mol)?
Depression of freezing point in any dilute solution is directly proportional to ______
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.
Of the following four aqueous solutions, total number of those solutions whose freezing points is lower than that of 0.10 M C2H5OH is ______. (Integer answer)
- 0.10 M Ba3 (PO4)2
- 0.10 M Na2 SO4
- 0.10 M KCl
- 0.10 M Li3 PO4
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 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 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.
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.
