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Question
A 5% aqueous solution (by mass) of cane sugar (molar mass 342 g mol-1) has a freezing point of 271 K. Calculate the freezing point of 5% aqueous glucose solution.
Solution
Given: Percentage by mass of cane sugar solution = 5 %
Percentage by mass of glucose solution = 5 %,
Freezing point of cane sugar solution = 271 K
Molar mass of cane sugar = 342 g mol-1
To find: Freezing point of glucose solution
Formula: `"M"_2 = (1000 xx "K"_"f" xx "W"_2)/(triangle "T"_"f" "W"_1)`
Calculation: 5 % solution (by mass) of cane sugar means that mass of cane sugar (W2) = 5 g, and mass of solvent (W1) = 95 g.
5 % glucose solution means that mass of glucose = `"W"_2^'` = 5g, and mass of solvent = `"W"_1^'` = 95 g
Molar mass of glucose (C6H12O6) = `("M"_2^')` = 180 g mol-1
Δ Tf for cane sugar solution = `triangle "T"_"f" = "T"_"f"^0 - "T"_"f"` = 273.15 K - 271 K = 2.15 K
Now, using the formula,
`"M"_2 = (1000 xx "K"_"f" xx "W"_2)/(triangle "T"_"f" "W"_1)`
Rearranging the formula, we get
`1000 "K"_"f" = ("M"_2 xx triangle "T"_"f" xx "W"_1)/"W"_2` ......(1)
`1000 "K"_"f" = ("M"_2^' xx triangle "T"_"f"^' xx "W"_1^')/"W"_2^'` .....(2)
From equations (1) and (2),
`("M"_2 xx triangle "T"_"f" xx "W"_1)/"W"_2 = ("M"_2^' xx triangle "T"_"f"^' xx "W"_1^')/"W"_2^'`
∴ `(342 "mol"^-1 xx 2.15 "K" xx 95 "g")/"5 g" = (180 "g mol"^-1 xx triangle "T"_"f"^' xx 95 "g")/"5 g"`
`triangle "T"_"f"^' = (342 "mol"^-1 xx 2.15 "K")/(180 "g mol"^-1)` = 4.085 K
∴ Freezing point of glucose solution (Tf) = `"T"_"f"^0 - triangle "T"_"f"^'`
= 273.15 K - 4.085 K
= 269.065 K
Freezing point of glucose solution is 269.065 K.
Alternate method:
Formulae: 1. m = `(1000 "W"_2)/("M"_2 "W"_1)`
2. Δ Tf = Kfm
Calculation: 5 % solution (by mass) of cane sugar means that mass of cane sugar (W2) = 5 g, and mass of solvent (W1) = 95 g.
Now, using formula (i),
Molality of cane sugar solution = m = `(1000 "W"_2)/("M"_2 "W"_1)`
`= (1000 "g kg"^-1 xx 5 "g")/(342 "g mol"^-1 xx 95 "g")`= 0.1539 m
Now, Δ Tf for cane sugar solution = Δ Tf = `triangle "T"_"f"^0 - "T"_"f"` = 273.15 K - 271 K = 2.15 K
From formula (ii),
Δ Tf (cane sugar) = Kf × m
`"K"_"f" = (triangle "T"_"f")/"m"`
∴ `"K"_"f" = 2.15/0.1539` = 13.97 K kg mol-1
5 % glucose solution means that mass of glucose = `"W"_2^'` = 5 g,
and mass of solvent = `"W"_1^'` = 95 g
Molar mass of glucose (C6H12O6) = `("M"_2^')` = 180 g mol-1
Using formula (i),
Molality of glucose solution =
m = `(1000 "W"_2)/("M"_2 "W"_1)`
= `(1000 "g kg"^-1 xx 5 "g")/(180 "g mol"^-1 xx 95 "g")` = 0.2924 m
From formula (ii),
`triangle "T"_"f"^' ("glucose") = "K"_"f" xx "m"`
(⸪ Since solvent is same, Kf is same for cane sugar and glucose solutions.)
∴ `triangle "T"_"f"^' ("glucose") = 13.97 xx 0.2924 = 4.085 "K"`
∴ Freezing point of glucose solution `("T"_"f") = "T"_"f"^0 - triangle "T"_"f"^'`
= 273.15 K - 4.085 K
= 269.065 K
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