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CHEM 3411, Fall 2010
Solution Set 5
In this solution set, an underline is used to show the last significant digit of numbers. For instance in x = 2.51693 the 2,5,1, and 6 are all significant. Digits to the right of the underlined digit, the 9 & 3 in the example, are not significant and would be rounded off at the end of calculations. Carrying these extra digits for intermediate values in calculations reduces rounding errors and ensures we get the same answer regardless of the order of arithmetic steps.
Numbers without underlines (including final answers) are shown with the proper number of sig figs.
1
Exercise 5.5b pg 201
Given
The vapour pressure of 2-propanol (Mw,p = 60.1 g/ mol) is 50.00 kPa at 338.8◦ C, but it fell to 49.62 kPa when mu = 8.69 g of an involatile organic compound was dissolved in mp = 250 g of 2-propanol.
In terms of given variables, this is written:
Mw,p = 60.1 g/ mol
T = 338.8◦ C p∗ = 50.0 kPa p pp = 49.62 kPa
Pressure of pure 2-propanol solvent
Pressure of solution containing unknown
mu = 8.69 g mp = 250 g
Find
Calculate the molar mass Mw,u of this unkown compound.
Strategy
We’ll start with Raoult’s Law (eq 5.21 on pg 164) which relates the partial pressure pa of a liquid in a solution to its pure vapour pressure p∗ by its mole fraction in solution χa . a pa = χa p∗ a This allows us to calculate the mole fraction of 2-propanol χp (in the solution with the dissolved unknown) to the solutions vapour pressure pp and the vapour pressure of pure 2-propanol p∗ . p χp = pp /p∗ p 1
CHEM 3411, Fall 2010
Solution Set 5
Additionally, we know the solution only contains 2-propanol and the unknown with moles fraction of χp and χu respectively, and therefore χp + χu = 1. Combining this with our expression for χp gives and expression for χu in terms of measured pressures. χu = 1 − pp /p∗ p Next, we’ll use the definition of mole fraction to relate χu to the moles of unknown by χu =
nu nu + np
Solution Set 5
In this solution set, an underline is used to show the last significant digit of numbers. For instance in x = 2.51693 the 2,5,1, and 6 are all significant. Digits to the right of the underlined digit, the 9 & 3 in the example, are not significant and would be rounded off at the end of calculations. Carrying these extra digits for intermediate values in calculations reduces rounding errors and ensures we get the same answer regardless of the order of arithmetic steps.
Numbers without underlines (including final answers) are shown with the proper number of sig figs.
1
Exercise 5.5b pg 201
Given
The vapour pressure of 2-propanol (Mw,p = 60.1 g/ mol) is 50.00 kPa at 338.8◦ C, but it fell to 49.62 kPa when mu = 8.69 g of an involatile organic compound was dissolved in mp = 250 g of 2-propanol.
In terms of given variables, this is written:
Mw,p = 60.1 g/ mol
T = 338.8◦ C p∗ = 50.0 kPa p pp = 49.62 kPa
Pressure of pure 2-propanol solvent
Pressure of solution containing unknown
mu = 8.69 g mp = 250 g
Find
Calculate the molar mass Mw,u of this unkown compound.
Strategy
We’ll start with Raoult’s Law (eq 5.21 on pg 164) which relates the partial pressure pa of a liquid in a solution to its pure vapour pressure p∗ by its mole fraction in solution χa . a pa = χa p∗ a This allows us to calculate the mole fraction of 2-propanol χp (in the solution with the dissolved unknown) to the solutions vapour pressure pp and the vapour pressure of pure 2-propanol p∗ . p χp = pp /p∗ p 1
CHEM 3411, Fall 2010
Solution Set 5
Additionally, we know the solution only contains 2-propanol and the unknown with moles fraction of χp and χu respectively, and therefore χp + χu = 1. Combining this with our expression for χp gives and expression for χu in terms of measured pressures. χu = 1 − pp /p∗ p Next, we’ll use the definition of mole fraction to relate χu to the moles of unknown by χu =
nu nu + np