Advertisements
Advertisements
Question
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.
Solution
Here, ΔTf = (273.15 − 271) K
= 2.15 K
Molar mass of sugar (C12H22O11) = 12 × 12 + 22 × 1 + 11 × 16
= 342 g mol−1
5% solution (by mass) of cane sugar in water means 5 g of cane sugar is present in (100 − 5)g = 95 g of water.
Now, number of moles of cane sugar = `5/342` mol
= 0.0146 mol
Therefore, molality of the solution, m = `(0.0146 "mol")/(0.095 "kg")`
= 0.1537 mol kg−1
Applying the relation,
ΔTf = Kf × m
`=> "K"_"f" = (triangle"T"_"f")/"m"`
`=> "K"_"f" = (2.15 "K")/(0.1537 "mol kg"^(-1))`
Kf = 13.99 K kg mol−1
Molar of glucose (C6H12O6) = 6 × 12 + 12 × 1 + 6 × 16
= 180 g mol−1
5% glucose in water means 5 g of glucose is present in (100 − 5) g = 95 g of water.
Number of moles of glucose = `5/180` mol
= 0.0278 mol
Therefore, molality of the solution, m = `(0.0278 "mol")/(0.095 "kg")`
= 0.2926 mol kg−1
Applying the relation,
ΔTf = Kf × m
= 13.99 K kg mol−1 × 0.2926 mol kg−1
= 4.09 K (approximately)
Hence, the freezing point of the 5% glucose solution is (273.15 − 4.09) K = 269.06 K.
APPEARS IN
RELATED QUESTIONS
Write the formula to determine the molar mass of a solute using freezing point depresssion method.
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.
Define Freezing point.
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 .)
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:
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?
The freezing point of equimolal aqueous solution will be highest for ____________.
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?
How does sprinkling of salt help in clearing the snow covered roads in hilly areas? Explain the phenomenon involved in the process.
Latent heat of water (ice) is 1436.3 cal per mol. What will be molal freezing point depression constant of water (R = 2 cal/degree/mol)?
Depression of freezing point in any dilute solution is directly proportional to ______
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 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.
