32 g of methane (molar mass 16 g/mol) is mixed with 192 g of oxygen (molar mass 32 g/mol). Presuming that these gases donot react with each other, what is the mole fraction of methane ?

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Correct — option (1), 0.25. The calculation has three steps and no shortcuts are needed. First convert each mass to a number of moles by dividing by the molar mass the question itself supplies. Methane: 32 grams divided by 16 grams per mole gives 2 moles. Oxygen: 192 grams divided by 32 grams per mole gives 6 moles. Second, add them to get the total amount of gas present, 2 plus 6, which is 8 moles. Third, apply the definition of mole fraction — the number of moles of the component of interest divided by the total number of moles of all components present. For methane that is 2 divided by 8, which is 0.25. As a check, the mole fraction of oxygen is 6 divided by 8, or 0.75, and the two add to exactly 1, as the mole fractions of all the components of any mixture must. Three features of the answer are worth noticing because they let a candidate reject wrong options without redoing the arithmetic. A mole fraction is a ratio of one amount to a larger amount, so it can never be greater than 1 and can never be negative; any option that is a whole number bigger than one is disqualified on sight. A mole fraction has no unit, because moles have been divided by moles, which is exactly why it is preferred to concentration when a quantity has to be independent of temperature and volume. And the ratio is by amount of substance, not by mass: 192 grams of oxygen weighs six times as much as 32 grams of methane, but it contains only three times as many molecules, because an oxygen molecule is twice as heavy as a methane molecule. Confusing mass with amount is the single most common error in this calculation and it accounts for one of the offered figures directly. The stem's condition that the two gases do not react — printed as 'donot' in the English column — is not decoration. Methane and oxygen react readily on ignition, one molecule of methane consuming two of oxygen to give carbon dioxide and water, and if the reaction were allowed to proceed the amounts of each gas present would change and the question would have no single answer. By stipulating that they are merely mixed, the examiner freezes the composition at the values calculated above.

Why the other options are wrong — (b) Six is the number of moles of oxygen in the mixture, obtained by dividing 192 grams by the molar mass of 32 grams per mole. It is a real quantity, and a necessary step on the way to the answer, but it is an amount and not a fraction, and it is the amount of the wrong component. It can be rejected without any arithmetic at all: a mole fraction is a part divided by the whole, so it always lies between 0 and 1, and no mixture can have a mole fraction of six. Options that offer an intermediate result of the calculation as though it were the final answer are a standard device in numerical questions, and the guard against them is to finish by checking that the number has the right form — a mole fraction between zero and one and no unit attached.  (c) This figure comes from dividing the two masses directly — 32 grams of methane by 192 grams of oxygen, which is one sixth, or 0.167 — and so from skipping the conversion to moles altogether. That is the error the question is built to catch. Equal masses of two gases do not contain equal numbers of molecules unless the molar masses happen to be equal, and here they are not: the molar mass of oxygen, 32, is twice that of methane, 16, so a given mass of oxygen contains only half as many molecules as the same mass of methane. Working in moles gives 2 and 6, a ratio of one to three; working in grams gives 32 and 192, a ratio of one to six. Notice too that 0.167 is not even the right form of answer if the masses had been the right quantities to use, since a mole fraction is taken over the total and not over the other component alone.  (d) Four is the answer turned upside down: it is the total amount of gas, 8 moles, divided by the amount of methane, 2 moles, instead of the amount of methane divided by the total. The mistake is one of definition rather than of arithmetic, and it is worth fixing the definition firmly for that reason — the mole fraction of a component is the moles of that component over the moles of everything present, so the component's own quantity is always on top and the total always below. Note how the option set has been assembled: 0.25 and 4 are reciprocals of each other, and so are 0.167 and 6, so each of the two ways of misreading the ratio has been printed alongside its inverted form. A candidate who has the definition right and remembers that the value must lie between zero and one is left with only one candidate figure.

Concept: The mole is the chemist's unit of amount of substance: one mole of any substance contains the Avogadro number of particles, about 6.022 times ten to the twenty-third, and has a mass in grams equal to its relative molecular or atomic mass. So one mole of methane weighs 16 grams and one mole of oxygen molecules weighs 32 grams, and converting between mass and amount is always a matter of dividing the mass by the molar mass. Composition of a mixture can then be expressed in several ways, and the examinable point is that they are not interchangeable. Mole fraction is the moles of one component divided by the total moles of all components; it is dimensionless, lies between 0 and 1, and the mole fractions of all components of a mixture sum to 1. Mass fraction, or percentage by mass, uses masses in the same way and gives a different number whenever the molar masses differ. Molarity is moles of solute per litre of solution and depends on temperature because volume does; molality is moles of solute per kilogram of solvent and does not. Mole fraction earns its place in two standard laws. Dalton's law of partial pressures states that in a mixture of gases the partial pressure of a component equals its mole fraction multiplied by the total pressure, so in the mixture of this question methane would exert one quarter of the total pressure and oxygen three quarters. Raoult's law states that the vapour pressure of a component of an ideal solution is its mole fraction multiplied by the vapour pressure of the pure component, and the colligative properties — lowering of vapour pressure, elevation of boiling point, depression of freezing point and osmotic pressure — all follow from it.

State PCS 2021 Q66 — Science & Technology PYQ with Answer

State PCS 2021 Q66 — Science & Technology PYQ with Answer