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प्रश्न
How does sprinkling of salt help in clearing the snow covered roads in hilly areas? Explain the phenomenon involved in the process.
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उत्तर
When salt is spread over snow covered roads, it lowers the freezing point of water to such an extent that water does not freeze to form ice. As a result, the snow starts melting from the surface and therefore, it helps in clearing the roads. Hence, common salt acts as de-icing agent.
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संबंधित प्रश्न
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
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.
Two elements A and B form compounds having formula AB2 and AB4. When dissolved in 20 g of benzene (C6H6), 1 g of AB2 lowers the freezing point by 2.3 K whereas 1.0 g of AB4 lowers it by 1.3 K. The molar depression constant for benzene is 5.1 K kg mol−1. Calculate the atomic masses of A and B.
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.
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.
Define Freezing point.
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)
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.
In comparison to a 0.01 m solution of glucose, the depression in freezing point of a 0.01 m MgCl2 solution is ______.
Which of the following statement is false?
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:
Which of the following statements is false?
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
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]
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 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.
