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Some bottles of colourless liquids were being labelled when the technicians accidentally mixed them up and lost track of their contents. A \(15.0 \mathrm{ml}\) sample withdrawn from one bottle weighed \(22.3 \mathrm{~g} .\) The technicians knew that the liquid was either acetone, benzene, chloroform on carbon tetrachloride(which have densities of \(0.792 \mathrm{~g} / \mathrm{cm}^{3}, 0.899 \mathrm{~g} / \mathrm{cm}^{3}, 1.489 \mathrm{~g} / \mathrm{cm}^{3}\), and \(1.595 \mathrm{~g} / \mathrm{cm}^{3}\), respectively). What was the identity of the liquid? (a) Carbon tetrachloride (b) Acetone (c) Chloroform (d) Benzene

Short Answer

Expert verified
The liquid is chloroform.

Step by step solution

01

Calculate the density of the sample

To determine the identity of the liquid, first calculate its density by using the mass (\(22.3 \text{ g}\)) and volume (\(15.0 \text{ ml}\)) of the sample. Density \(\rho\) is mass divided by volume, so \(\rho = \frac{m}{V}\). Note that \(1 \text{ ml} = 1 \text{ cm}^3\).
02

Determine the identity of the liquid

Compare the calculated density with the given densities of the possible liquids to determine which liquid's density is closest to the one calculated. The liquid whose density matches the calculated density is the identity of the sample.

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Key Concepts

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

Physical Chemistry
Physical chemistry is a branch of chemistry that deals with understanding the physical properties of molecules, the forces that act upon them, and their reactions based on these properties. It blends the principles of physics and chemistry to study the physical characteristics of molecules, their dynamic behaviour, and how they interact with energy.

One crucial aspect of physical chemistry is the study of density, a physical property that relates the mass of a substance to its volume. The concept of density (\rho) is essential in physical chemistry because it allows chemists to identify substances and understand their composition and purity. The formula for density is expressed as \rho = \frac{m}{V}\r, where \(m\) is mass and \(V\) is volume.

In the given exercise, the density determination serves as a fundamental approach to identify a colorless liquid based on its physical properties. By calculating the density of a substance, students can compare known densities of substances to identify unknown samples, a practical skill in physical chemistry.
Chemical Property Analysis
Chemical property analysis involves evaluating the characteristics of a substance that become evident during a chemical reaction. Unlike physical properties, which can be determined without changing the substance's composition, chemical properties describe a substance's potential to undergo specific chemical changes. Density, while often considered a physical property, is crucial in chemical property analysis as it can influence how substances interact during reactions.

For example, understanding the density of reactants can predict the separation of layers in immiscible liquids or the stratification of gases. Moreover, density plays a role in determining the purity of a substance, as impurities can significantly alter a substance's density. In the context of the exercise, the problem revolves around using the physical property of density to discriminate between different chemical substances, emphasizing the importance of chemical property analysis in practical scenarios.

Moreover, using density determination as a method for identifying substances is common in chemical property analysis in laboratory settings, as seen in this problem where technicians use it to identify a liquid by its mass-to-volume ratio.
Competitive Examinations Chemistry Problems
Competitive examinations often include chemistry problems that test a student's ability to apply concepts from physical and analytical chemistry to solve practical problems. These problems require a solid understanding of fundamental concepts, attention to detail, and the ability to think critically under time constraints.

Problems like the one presented here, involving the identification of an unknown liquid by calculating its density, are typical in competitive examinations. They test the student's grasp of formulae, units of measurement (recognizing that 1 ml is equivalent to 1 cm³), and their problem-solving skills.

To excel in such problems, students should practice converting units, applying concepts to different scenarios, and interpreting results accurately. They should also develop a strategic approach to eliminate incorrect options, as was done by calculating the density and methodically checking which known substance it matched with. This strategic approach not only saves time but also enables the test-taker to efficiently navigate through trickier sections of competitive chemistry examinations.

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Most popular questions from this chapter

The empirical formula of an organic gaseous compound containing carbon and hydrogen is \(\mathrm{CH}_{2}\). The volume occupied by certain mass of this gas is exactly half of the volume occupied by the same mass of nitrogen gas under identical conditions. The molecular formula of the organic gas is (a) \(\mathrm{C}_{2} \mathrm{H}_{4}\) (b) \(\mathrm{CH}_{2}\) (c) \(\mathrm{C}_{6} \mathrm{H}_{12}\) (d) \(\mathrm{C}_{4} \mathrm{H}_{8}\)

A sample of iron oxide has \(\mathrm{FeO}\) and \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) in the mole ratio \(2: 1\). It is partially oxidized to change this ratio to \(1: 2\). The number of moles of \(\mathrm{FeO}\) oxidized per mole of initial mixture is (a) \(0.2\) (b) \(0.333\) (c) \(0.4\) (d) \(0.5\)

A quantity of \(10 \mathrm{~g}\) of a hydrocarbon exactly requires \(40 \mathrm{~g}\) oxygen for complete combustion. The products formed are \(\mathrm{CO}_{2}\) and water. When \(\mathrm{CO}_{2}\) gas formed is absorbed completely in lime water, the mass of solution increases by \(27.5 \mathrm{~g}\). What is the mass of water formed in combustion? (a) \(22.5 \mathrm{~g}\) (b) \(27.5 \mathrm{~g}\) (c) \(50 \mathrm{~g}\) (d) \(10 \mathrm{~g}\)

A quantity of \(2.0 \mathrm{~g}\) of a triatomic gaseous element was found to occupy a volume of \(448 \mathrm{ml}\) at \(76 \mathrm{~cm}\) of \(\mathrm{Hg}\) and \(273 \mathrm{~K}\). The mass of its each atom is (a) 100 amu (b) \(5.53 \times 10^{-23} \mathrm{~g}\) (c) \(33.3 \mathrm{~g}\) (d) \(5.53\) amu

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) \(\frac{100}{110}\)

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