1) a) What is the normail boiling point (760 mm Hg) for a compound that boils at 150 degree celsius at 10 mm Hg pressure?
b) At which temperature would the compund in (a) boil if the pressure were 40 mm Hg?
c) A compound was distilled at atmospheric pressure and had a boiling point of 285 degree celsius.WHat would be the approximate boiling range for this compound at 15 mm Hg?
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a) To determine the normal boiling point of a compound, you can use the concept of boiling point elevation. The boiling point of a liquid increases as the pressure also increases. The normal boiling point is defined as the boiling point at a pressure of 760 mm Hg (also known as 1 atmosphere or atm).
In this case, we know that the compound boils at 150 degrees Celsius at a pressure of 10 mm Hg. We need to find the temperature at 760 mm Hg.
To calculate the normal boiling point, we can use the Clausius-Clapeyron equation:
ln(P1/P2) = (ΔHvap/R) * (1/T2 - 1/T1)
Where:
P1 and P2 are the pressures (10 mm Hg and 760 mm Hg)
T1 and T2 are the temperatures (150 degrees Celsius and the unknown boiling point at 760 mm Hg)
ΔHvap is the heat of vaporization for the compound
R is the ideal gas constant (8.314 J/mol∙K)
Since we don't have the heat of vaporization, we'll assume it to be constant and cancel it out in the equation. Therefore, the equation becomes:
ln(P1/P2) = (1/T2 - 1/T1)
Rearranging the equation and substituting the values, we have:
ln(10/760) = 1/T2 - 1/423.15
Simplifying further:
-3.21888 = 1/T2 - 0.00236
Now, solving for 1/T2:
1/T2 = -3.21888 + 0.00236
1/T2 = -3.21652
Finally, we can find T2:
T2 = 1 / (-3.21652)
T2 ≈ -0.3102 K
Since the temperature cannot be negative, it implies that the pressure given (10 mm Hg) is likely below the range of accuracy for the Clausius-Clapeyron equation, or we might have made an error in the calculations.