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Question
Which of the following techniques would allow for separation of a mixture of two liquids based on intermolecular interactions with mobile and stationary phases?
| A. Distillation | |
| B. Column chromatography | |
| C. Filtration | |
| D. Titration |
Explanation
Chromatography (e.g., column chromatography, thin-layer chromatography, or paper chromatography) is a technique that uses a stationary phase and a solvent (a mobile phase) to separate the components of a mixture according to their polarity. Separation occurs because the mixture components travel at different rates, depending on the strength of the competing intermolecular interactions between the mixture component and the stationary and mobile phases.
For example, when a mixture is separated by column chromatography using a polar stationary phase and a nonpolar mobile phase, a more polar compound will travel slowly down the column because it has stronger intermolecular interactions with the stationary phase than with the mobile phase. Conversely, a less polar compound will travel quickly down the column because it has stronger intermolecular interactions with the mobile phase than with the stationary phase.
Therefore, column chromatography can be used to separate a mixture of two liquids based on intermolecular interactions with a stationary and mobile phase.
(Choice A) Although distillation can be used to separate a mixture of two liquids, distillation does not have a mobile and a stationary phase. Instead, it is based on boiling points and intermolecular forces between the molecules of the mixture.
(Choice C) Filtration is a physical process used to separate a solid from a liquid, and cannot be used to separate a mixture of liquids.
(Choice D) Titration is a technique used for acid-base neutralization.
Things to remember:
Separation of a mixture using chromatography occurs because the mixture components travel through the column at different rates, depending on the strengths of the competing intermolecular interactions between each component and the stationary phase and mobile phase.
Passage
The three molecules shown in the table above are liquids at room temperature, and each have the molecular formula C5H10O. A 0.02 mole sample of each compound is separately injected into three identical containers, each with a movable piston. The compounds are vaporized by heating to 420 K, causing each piston to rise to keep the internal pressure equal to atmospheric pressure (1 atm), as illustrated below.
Question
What is the expected relationship between the volume V occupied by each gas and the gas constant R when the containers are at 420 K, based on the ideal gas law? Note: use units of (atm × L)/(mole × K) for R.
| A.
V =
1
4.2R
|
|
| B.
V =
1
8.4R
|
|
| C.
V =
4.2R
|
|
| D.
V =
8.4R
|
Explanation
The ideal gas law relates the pressure P and volume V of a gas to the number of moles n and the Kelvin temperature T of that gas by the equation PV=nRT, where R is the gas constant. When all parameters except one are known, the remaining parameter can be determined by solving the equation.
The passage states that the pressure inside the vessel is kept at 1 atm, the temperature is 420 K, and 0.02 moles of each compound were used. These values can be substituted into the ideal gas equation:
1 atm × V = 0.02moles × R × 420k
The value of V can then be solved in terms of R
This expression can then be simplified to yield the relationship of V to R:
V = 8.4 R
(Choice A) is the result of incorrectly dividing the pressure by nRT instead of dividing nRT by pressure, and using only 0.01 moles of gas instead of 0.02 moles.
(Choice B) is the result of incorrectly dividing pressure by nRT instead of dividing nRT by pressure. It is the reciprocal of the correct answer.
(Choice C) 4.2R is the result of mistakenly using only 0.01 moles of each compound instead of 0.02 moles.
Things to remember:
The ideal gas law (PV = nRT) relates the pressure and volume of an ideal gas to the temperature and number of moles of that gas. When all parameters but one are known, the remaining one can be solved for algebraically.
Question
In an experiment, a chemist isolates a pure sample of Ca3(PO4)2 from dried bones. In another experiment, the chemist isolates a pure sample of Ca3(PO4)2 from hydroxyapatite, a phosphate mineral. Compared to the mass percent of calcium in the Ca3(PO4)2 from the mineral, the mass percent of calcium in the Ca3(PO4)2 from the dried bones is
| A. lower, because some of the Ca3(PO4)2 is not transferred from the lab vessels during the experiment. | |
| B. lower, because dried bones contain less Ca3(PO4)2 than the mineral. | |
| C. the same, because the ratio of the masses of the constituent elements is the same in both samples. | |
| D. higher, because dried bones contain more Ca3(PO4)2 than the mineral. |
Explanation
The law of definite proportions states that a pure sample of a given compound always contains the same proportion (mass ratio) of its constituent elements. This law results from the fact that atoms of different elements assemble (form chemical bonds) in specific whole-number ratios.
For example, each formula unit of Ca3(PO4)2 always contains three Ca atoms, two P atoms, and eight O atoms (a 3:2:8 atom ratio). Each type of atom in a formula unit has a different size and mass that contributes a percentage of the total formula unit mass, which is found by summing the contributing atomic masses from the periodic table.
Calculating the mass percentage of each element gives:
Accordingly, any pure sample of Ca3(PO4)2 will always be 38.8% Ca, 20.0% P, and 41.3% O by mass. Dividing each of these percentages by the smallest value gives the mass ratio of these elements in Ca3(PO4)2.
Consequently, all pure samples of Ca3(PO4)2 will have a mass ratio of 1.94 parts calcium to 1.00 part phosphorus to 2.07 parts oxygen, regardless of the sample source.
Therefore, the mass percent of calcium in the Ca3(PO4)2 from the dried bones compared to the mass percent of calcium in the Ca3(PO4)2 from the mineral is the same.
(Choice A) Although transfer losses can reduce the total mass recovered, the mass percentages of the elements in the isolated Ca3(PO4)2 samples will be the same.
(Choices B and D) An impure sample may contain more (or less) of a given compound than another impure sample, but when a compound is isolated and purified, the mass percentages of the elements in the pure isolated compound will be the same.
Things to remember:
According to the law of definite proportions, a pure sample of a given compound always contains the same proportion (mass ratio) of its constituent elements.
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