Advertisements
Advertisements
Question
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)
Advertisements
Solution
Given,
W2 = 60 g
M2 = 180 g mol-1
W1 = 250 g
Kf = 1.86 K kg mol–1
ΔTf = kfm
`T_f^0 - T_f = (K_f xx w_2 xx 1000)/M_2xxw_1`
`2.73 - T_f = (1.86 xx 60 xx 1000)/(180xx250)`
`2.73 - T_f = (1.86 xx 60 xx 1000)/(180 xx 250)`
273.15 - Tf = 2.48
Tf = 273.15 - 2.48 = 270.67 K
Hence, the freezing point of water is 270.67 K or -2.48°C
APPEARS IN
RELATED QUESTIONS
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.
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)
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 ____________.
If molality of dilute solution is doubled, the value of molal depression constant (Kf) will be ______.
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 ______
Of the following four aqueous solutions, total number of those solutions whose freezing points is lower than that of 0.10 M C2H5OH is ______. (Integer answer)
- 0.10 M Ba3 (PO4)2
- 0.10 M Na2 SO4
- 0.10 M KCl
- 0.10 M Li3 PO4
Out of the following 1.0 M aqueous solution, which one will show the largest freezing point depression?
