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Question
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)
Solution
Molality (m) = `"n"/"W"_("solvent"("kg"))`
For sucrose solution:
m =`(4/342)/(96/1000)= 4/342xx1000/96 = 0.122"m"`
(Δ Tf)1 = (273.15 − 271.15)K = 2K
(Δ Tf)1 = Kfm = Kf × 0.122
2= Kf × 0.122 .......(1)
For glucose solution :
m=`(5/180)/(95/1000)=5/180xx1000/95= 0.292"m"`
(Δ Tf)2 =Kf × 0.292 .......(2)
Dividing eqn. (2) by (1)
`(Delta "T"_"f")_2/2 = ("K"_"f"xx0.292)/("K"_"f"xx0.122`
`(Delta"T"_"f")_2 = 0.292/0.122xx2 = 4.79`
Tf = 273.15 − 4.79 = 268.36 K
Freezing point of glucose solution will be 268.36 K
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| 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 |
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| 3 | 0.5 | -3.0 | -5.5 |
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