6- What is the freezing point of an aqueous solution that boils at 105.docx

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6. What is the freezing point of an aqueous solution that boils at 105.0 oC? Solution The boiling point of an aqueous solution is 105.0Â°C. The boiling point of pure water is 100.0Â°C. The elevation of boiling point is given as ?T b = (boiling point of solution) â€“ (boiling point of pure water) = (105.0Â°C) â€“ (100.0Â°C) = 5.0Â°C. As per Raoultâ€™s law, we have, ?T b = K b *(molality of the solution) where K b = 0.512Â°C/m is the boiling point elevation constant of water. Plug in values and obtain 5.0Â°C = (0.512Â°C/m)*(molality of the solution) ====> molality of the solution = (5.0Â°C)/(0.512Â°C/m) = 9.7656 m. Assume that the molality of the solution remains constant; therefore, we can find the freezing point depression of the aqueous solution using Raoultâ€™s law as below. ?T f = K f *(molality of the solution) where K f = freezing point depression constant of water = 1.86Â°C/m. Plug in values and get ?T f = (1.86Â°C/m)*(9.7656 m) = 18.1640Â°C. We define ?T f = (T f ) pure â€“ (T f ) solution where T f,pure = 0.0Â°C is the freezing point of pure water and T f,solution is the freezing point of the aqueous solution. Plug in values and get 18.1640Â°C = 0.0Â°C â€“ T f , solution ====> T f,solution = 0.0Â°C â€“ 18.1640Â°C = -18.1640Â°C ? -18.2Â°C (ans). .

6. What is the freezing point of an aqueous solution that boils at 105.0 oC?
Solution
The boiling point of an aqueous solution is 105.0Â°C. The boiling point of pure water is
100.0Â°C. The elevation of boiling point is given as
?T b = (boiling point of solution) â€“ (boiling point of pure water) = (105.0Â°C) â€“ (100.0Â°C)
= 5.0Â°C.
As per Raoultâ€™s law, we have,
?T b = K b *(molality of the solution)
where K b = 0.512Â°C/m is the boiling point elevation constant of water. Plug in values and
obtain
5.0Â°C = (0.512Â°C/m)*(molality of the solution)
====> molality of the solution = (5.0Â°C)/(0.512Â°C/m) = 9.7656 m.
Assume that the molality of the solution remains constant; therefore, we can find the freezing
point depression of the aqueous solution using Raoultâ€™s law as below.
?T f = K f *(molality of the solution)
where K f = freezing point depression constant of water = 1.86Â°C/m. Plug in values and get
?T f = (1.86Â°C/m)*(9.7656 m) = 18.1640Â°C.
We define ?T f = (T f ) pure â€“ (T f ) solution
where T f,pure = 0.0Â°C is the freezing point of pure water and T f,solution is the freezing point of the
aqueous solution. Plug in values and get
18.1640Â°C = 0.0Â°C â€“ T f , solution

====> T f,solution = 0.0Â°C â€“ 18.1640Â°C = -18.1640Â°C ? -18.2Â°C (ans).

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