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A given mixture consists only of pure substance \(X\) and pure substance \(Y\). The total mass of the mixture is \(3.72 \mathrm{~g}\). The total number of moles is \(0.06\). If the mass of one mole of \(Y\) is \(48 \mathrm{~g}\) and there is \(0.02\) mole of \(X\) in the mixture, what is the mass of one mole of \(\mathrm{X}\) ? (a) \(90 \mathrm{~g}\) (b) \(75 \mathrm{~g}\) (c) \(45 \mathrm{~g}\) (d) \(180 \mathrm{~g}\)

Short Answer

Expert verified
The mass of one mole of X is 90 g/mole.

Step by step solution

01

Calculate the total moles of Y

Since the total number of moles in the mixture is 0.06 and there is 0.02 mole of X, subtract the moles of X from the total moles to find the moles of Y: Moles of Y = Total moles - Moles of X = 0.06 moles - 0.02 moles = 0.04 moles
02

Calculate the mass of Y in the mixture

Multiply the moles of Y by the mass of one mole of Y to find the mass of Y in the mixture: Mass of Y = Moles of Y * Molar mass of Y = 0.04 moles * 48 g/mole = 1.92 g
03

Calculate the mass of X in the mixture

Subtract the mass of Y from the total mass of the mixture to find the mass of X: Mass of X = Total mass - Mass of Y = 3.72 g - 1.92 g = 1.80 g
04

Calculate the molar mass of X

Divide the mass of X by the moles of X to find the molar mass of X: Molar mass of X = Mass of X / Moles of X = 1.80 g / 0.02 moles = 90 g/mole

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Stoichiometry
Stoichiometry is like a recipe for chemistry. It is the calculation of reactants and products in chemical reactions. In other words, stoichiometry helps us figure out how much of each substance we need to start with and what we can expect to end up with at the end of a chemical reaction. It's all about the balance—the 'art' of making sure atoms are conserved during reactions and keeping track of the masses before and after the reaction occurs.

Imagine baking cookies; stoichiometry tells you how much flour, sugar, or chocolate chips are needed to make a certain number of cookies. Similarly, in chemistry, if you know the amount of one substance, it helps you determine the amounts of other substances that will react or be produced. In the exercise we're discussing, we use stoichiometric calculations to find out the balance between substance X and Y in a mixture.
Chemical Mixture
A chemical mixture contains more than one substance, and they're mixed together without any chemical bonding. Such mixtures can be separated into their pure components by physical methods, like filtration or distillation. Each component in a mixture retains its own chemical properties. Think of it as a smoothie, where you can blend strawberries and bananas; they mix, but each fruit's unique flavor still stands out.

In our exercise, we are dealing with a mixture of two substances, X and Y. The challenge here is to figure out how much of each substance is present. Understanding the concept of mixtures is crucial, as it emphasizes that even though X and Y are mixed, they still retain their individual identities (mass and molar mass), which allows us to calculate them separately.
Mole Concept
The mole concept is a bridge between the microscopic world of atoms and the macroscopic world we live in. It allows chemists to count particles in a given substance by weighing it. One mole of any substance contains Avogadro's number of particles, which is approximately \(6.022 \times 10^{23}\) particles.

In simple terms, the mole is a unit used to measure the amount of substance. Just like a dozen implies 12 items, a mole implies an incredibly large number of particles. This concept is vital when you want to figure out how many particles you have just by knowing the mass. In the exercise, the mole concept is used to convert mass of substances to moles and vice versa, providing an understanding of how much of each substance is present in the mixture based on the molar mass and the number of moles.

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Most popular questions from this chapter

A quantity of \(500 \mathrm{~g}\) of a urea solution of mole fraction \(0.2\) is diluted to \(1500 \mathrm{~g}\). The mole fraction of solute in the diluted solution is (a) \(0.05\) (b) \(0.067\) (c) \(0.6\) (d) \(0.1\)

The volume of one mole of water at \(277 \mathrm{~K}\) is \(18 \mathrm{ml}\). One \(\mathrm{ml}\) of water contains 20 drops. The number of molecules in one drop of water will be \(\left(N_{\mathrm{A}}=6 \times 10^{23}\right)\) (a) \(1.07 \times 10^{21}\) (b) \(1.67 \times 10^{21}\) (c) \(2.67 \times 10^{21}\) (d) \(1.67 \times 10^{20}\)

An aqueous solution has urea and glucose in mass ratio \(3: 1\). If the mass ratio of water and glucose in the solution is \(10: 1\), then the mole fraction of glucose in the solution is (a) \(\frac{1}{110}\) (b) \(\frac{9}{110}\) (c) \(\frac{3}{110}\) (d) \(\frac{100}{110}\)

The density of a DNA sample is \(1.1 \mathrm{~g} / \mathrm{ml}\) and its molar mass determined by cryoscopic method was found to be \(6 \times 10^{8} \mathrm{~g} / \mathrm{mole}\). What is the volume occupied by one DNA molecule? \(\left(N_{\mathrm{A}}=6 \times 10^{23}\right)\) (a) \(5.45 \times 10^{8} \mathrm{ml}\) (b) \(1.83 \times 10^{-9} \mathrm{ml}\) (c) \(9.06 \times 10^{-16} \mathrm{ml}\) (d) \(1.09 \times 10^{-13} \mathrm{ml}\)

Samples of \(1.0 \mathrm{~g}\) of \(\mathrm{Al}\) are treated separately with an excess of sulphuric acid and an excess of sodium hydroxide. The ratio of the number of moles of the hydrogen gas evolved is (a) \(1: 1\) (b) \(3: 2\) (c) \(2: 1\) (d) \(9: 4\)

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