Basis of Calculation and Composition Questions and Answers

Answer: C

One milligram in one kilogram is one part in a million by mass, which is the working definition. For dilute aqueous solutions it is nearly the same as milligrams per litre, since the density is close to one kilogram per litre.

Answer: D

Correct answer: the specific gravity of the solution

Specific gravity is a bulk physical property - a ratio of densities - and says nothing directly about what the solution contains; two quite different mixtures may share one. Mole fraction, mass fraction and molality all state how much of a species is present relative to a defined quantity, which is what a composition unit does. Specific gravity is used to *infer* composition only through a calibration for one particular system.

Answer: A

The gas is passed through absorbents over water, so the water vapour condenses and is not counted. Converting an Orsat analysis to a wet basis requires the water to be found from a hydrogen balance. Reporting the basis with the analysis avoids the ambiguity.

Answer: B

Both are true and the reason gives the arithmetic: the same moles of carbon dioxide divided by a smaller total give a larger fraction, so the basis must always be stated with the figure. Dry basis figures are the ones usually quoted for flue gas.

Answer: C

Removing the water leaves eighty moles per hundred, and ten divided by eighty is twelve and a half per cent. Dividing by the original hundred instead is the standard mistake. Subtracting the water from the total comes first.

Answer: D

Equal volumes of ideal gases at the same conditions contain equal numbers of moles, so the two measures coincide. The equality fails for liquids, where volumes are not additive on mixing. Gas mixtures are the only case where the two coincide exactly.

Answer: A

Mixing ethanol and water gives a total volume less than the sum of the parts, so a volume per cent does not describe a definite mixture. Mass and mole bases have no such difficulty. Ethanol and water are the standard demonstration of the effect.

Answer: B

Molarity uses a volume of solution, which expands with temperature. The other three are all ratios of masses or mole numbers and are fixed once the mixture is made. Any unit built on a volume inherits the same problem.

Answer: C

Multiplying each flow rate by one hour turns a continuous problem into a fixed set of quantities. A batch problem already provides its own quantities and needs no such conversion. The choice of one hour is arbitrary and any interval will do.

Answer: D

The first two are correct and are the practical rules. The third is false, since a change of basis is only a change of scale and any ratio in the answer is unaffected. Only the intermediate numbers differ between two bases.