हिंदी
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 ______.

विकल्प

  • the same

  • about twice

  • about three times

  • about six times

MCQ
रिक्त स्थान भरें
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
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 2: Solutions - Exercises [पृष्ठ १९]

APPEARS IN

एनसीईआरटी एक्झांप्लर Chemistry Exemplar [English] Class 12
अध्याय 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.


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 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.


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.


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 .) 


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:


The freezing point of equimolal aqueous solution will be highest for ____________.


If molality of dilute solution is doubled, the value of molal depression constant (Kf) will be ______.


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:

  1. One temperature in the second set of results does not fit the pattern. Which temperature is that? Justify your answer.
  2. Why did Henna collect two sets of results?
  3. 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.

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]


Ibrahim collected 10 mL each of fresh water and ocean water. He observed that one sample labeled “P” froze at 0° C while the other “Q” at -1.3° C. Ibrahim forgot which of the two, “P” or “Q” was ocean water. Help him identify which container contains ocean water, giving rationalization for your answer.


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 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.


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×