Advertisements
Advertisements
प्रश्न
Calculate the freezing point of solution when 1.9 g of MgCl2 (M = 95 g mol−1) was dissolved in 50 g of water, assuming MgCl2 undergoes complete ionization.
(Kf for water = 1.86 K kg mol−1)
Advertisements
उत्तर
Given:
Kf = 1.86 K kg mol−1
Mass of solute = 1.9 g
Mass of solvent = 50 g
Therefore,
`"Molality of the solution, "m=1.9/95xx1000/50=0.4m`
Also, MgCl2 undergoes complete ionisation and thereby yielding 3 moles of constituent ions for every mole of MgCl2.
∴ i = 3
Now, depression in freezing point is given as
∆Tf = iKfm
= 3×1.86×0.4
= 2.232 K
Tf = 273.15−2.232
=270.918 K
Hence, the new freezing point of the solution is 270.92 K.
APPEARS IN
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 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)
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:
Which of the following 0.10 m aqueous solutions will have the lowest 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.
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 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 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.
