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.

Answer 1

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 T_f 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 (T_f) = `"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 T_f = K_fm

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 T_f for cane sugar solution = Δ T_f = `triangle "T"_"f"^0 - "T"_"f"` = 273.15 K - 271 K = 2.15 K

From formula (ii),

Δ T_f (cane sugar) = K_f × 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