मराठी
कर्नाटक बोर्ड पी.यू.सी.पीयूसी विज्ञान 2nd PUC Class 12

In comparison to a 0.01 m solution of glucose, the depression in freezing point of a 0.01 m MgCl2 solution is ______.

Advertisements
Advertisements

प्रश्न

In comparison to a 0.01 m solution of glucose, the depression in freezing point of a 0.01 m MgCl2 solution is ______.

पर्याय

  • the same

  • about twice

  • about three times

  • about six times

MCQ
रिकाम्या जागा भरा
Advertisements

उत्तर

In comparison to a 0.01 m solution of glucose, the depression in freezing point of a 0.01 m MgCl2 solution is about three times.

Explanation:

0.01 m solution of glucose does not ionize while 0.01 m  MgCl2 solution furnishes 3 ions \[\ce{(Mg^{2+} + 2Cl^-)}\] in the solution, hence the value of colligative property for  MgCl2 solution is about 3 times.

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 2: Solutions - Exercises [पृष्ठ १९]

APPEARS IN

एनसीईआरटी एक्झांप्लर Chemistry Exemplar [English] Class 12
पाठ 2 Solutions
Exercises | Q I. 11. | पृष्ठ १९

संबंधित प्रश्‍न

Write the formula to determine the molar mass of a solute using freezing point depresssion method.


Define Cryoscopic constant.


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.


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.


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, K= 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) 


Cryoscopic constant of a liquid is ____________.


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 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:


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.


Calculate freezing point depression expected for 0.0711 m aqueous solution of Na2SO4. If this solution actually freezes at – 0.320°C, what will be the value of van't Hoff factor? (kg for water = 108b°C 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)


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.


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.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×