40 milligram diatomic volatile substance \(\left(X_{2}\right)\) is converted to vapour that displaced \(4.92 \mathrm{~mL}\) of air at 1 atm and \(300 \mathrm{~K}\). Atomic weight of element \(X\) is nearly: (a) 400 (b) 240 (c) 200 (d) 100

Short Answer

Expert verified
The atomic weight of element X is nearly 100.

Step by step solution

01

Convert Volume to Moles

Use the ideal gas law, which states that \(PV = nRT\), to convert the volume of gas produced to moles. Given: \(P = 1 \text{atm}\), \(V = 4.92 \text{mL}\), \(T = 300 \text{K}\), and \(R = 0.0821 \text{L atm} \text{mol}^{-1} \text{K}^{-1}\). First, convert the volume from milliliters to liters by dividing by 1000: \(V = 4.92 \text{mL} = 0.00492 \text{L}\). Now, rearrange the ideal gas law to solve for n (moles): \(n = \frac{PV}{RT}\).
02

Calculate Moles

Plug the values into the ideal gas law equation to find the number of moles. \(n = \frac{(1)(0.00492)}{(0.0821)(300)}\) which simplifies to \(n = \frac{0.00492}{24.63}\).
03

Calculate Molar Mass of Diatomic Molecule

Given that 40 milligrams (0.040 grams) of the substance yields n moles, we can find the molar mass (M) by dividing the mass by the number of moles: \(M = \frac{0.040 \text{g}}{n}\).
04

Calculate Atomic Weight of Element X

Since the substance is diatomic (\(X_2\)), the atomic weight of X will be half of the molar mass of \(X_2\). Calculate \(M\) from Step 3 and divide by 2 to find the atomic weight of X.
05

Evaluate the Atomic Weight of Element X

Once you have the molar mass of the diatomic molecule, substituting the value of n from Step 2 into the equation from Step 3: \(M = \frac{0.040}{0.00492/24.63}\), and after calculating M, divide it by 2 to find the approximate atomic weight. If the atomic weight is close to one of the answer choices, that would be the estimated atomic weight of the element X.

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Key Concepts

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

Ideal Gas Law
Understanding the behavior of gases is crucial in chemistry, and the Ideal Gas Law provides a practical tool for relating the pressure (P), volume (V), temperature (T), and number of moles (n) of an ideal gas. The law is summarized by the equation
\(PV = nRT\),
where R is the ideal gas constant. An 'ideal' gas is a hypothetical gas whose molecules have no intermolecular forces and occupy no volume. Although real gases do not perfectly adhere to these assumptions, the Ideal Gas Law offers a close approximation for many conditions.

In solving problems that involve the Ideal Gas Law, students must ensure that their units are consistent. For example, pressure should be in atmospheres (atm), volume in liters (L), and temperature in Kelvin (K). When given other units like milliliters or Celsius, conversions are necessary:
  • 1 mL = 0.001 L
  • Degrees Celsius to Kelvin = Degrees Celsius + 273.15
These conversions are essential to avoid any discrepancies in calculations. The Ideal Gas Law not only helps to calculate the number of moles of a gas given PVT conditions but also lays the ground for further calculations, such as determining the molar mass of a substance.
Molar Mass Determination
Molar mass is a fundamental concept in stoichiometry that is defined as the mass of one mole of a substance. It is typically measured in grams per mole (g/mol) and is vital for converting between mass and moles in chemical equations. The determination of molar mass becomes particularly significant when dealing with reactions involving gases, as in the original exercise.

To determine the molar mass of a gas, analytical methods such as mass spectrometry may be used; however, the molar mass can also be calculated from experimental data using the Ideal Gas Law. In the given problem, after finding the number of moles (n) of a gas using the Ideal Gas Law, the molar mass (M) can be calculated by dividing the mass of the gas sample by the number of moles:
\(M = \frac{\text{mass of the gas sample (g)}}{n}\).
If the substance is a compound, such as the diatomic molecule \(X_2\) mentioned in the exercise, this gives us the molar mass of the entire molecule. To find the atomic weight of a single atom within the molecule, further simple arithmetic is required.
Stoichiometry
Stoichiometry is the study of quantitative relationships in chemical reactions, often described as the 'mathematics of chemistry.' It is based on the laws of conservation of mass and the fixed composition of compounds. Stoichiometric calculations enable chemists to predict the amounts of reactants and products involved in a reaction.

The basic principle of stoichiometry is that in a chemical reaction, atoms are neither created nor destroyed, they are merely rearranged. Consequently, the amount of each element must be the same in the reactants and products. Therefore, knowing the molar mass of the substances involved is crucial for stoichiometry since it allows us to relate mass to moles and vice versa. For instance, to find the stoichiometric ratios in a balanced chemical equation, it's often helpful to convert all quantities to moles.

In the context of the original exercise, stoichiometry comes into play when translating the mass of a diatomic volatile substance (given in milligrams) into moles through the Ideal Gas Law and then into the atomic weight of the element (X). This process illustrates the interconnectedness of stoichiometry with other core chemical concepts.
Chemical Problem Solving
Tackling chemical problems like calculating an element's atomic weight from experimental data is a multi-step process that involves understanding and applying chemical principles and equations. The steps typically begin with converting units, applying laws such as the Ideal Gas Law for gases, and performing stoichiometric calculations.

Effective chemical problem-solving requires a systematic approach:
  • Clearly define what is known from the problem and what needs to be determined.
  • Identify the correct chemical principles and equations to use.
  • Perform unit conversions where necessary.
  • Carefully execute the calculations step by step.
  • Check the answer for consistency with the physical meaning and units.
In the given exercise, such an approach helps to seamlessly navigate from the volume of a gas to its moles, then to its molar mass, and finally to the atomic weight of the component element. This methodology is not only useful in solving homework problems but also crucial in laboratory settings and real-world applications.

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Most popular questions from this chapter

The chief ore of \(\mathrm{Zn}\) is the sulphide, \(\mathrm{ZnS}\). The ore is concentrated by froth floatation process and then heated in air to convert \(\mathrm{ZnS}\) to \(\mathrm{ZnO}\). $$ \begin{aligned} &2 \mathrm{ZnS}+3 \mathrm{O}_{2} \stackrel{80 \%}{\longrightarrow} 2 \mathrm{ZnO}+2 \mathrm{SO}_{2} \\ &\mathrm{ZnO}+\mathrm{H}_{2} \mathrm{SO}_{4} \stackrel{100 \%}{\longrightarrow} \mathrm{ZnSO}_{4}+\mathrm{H}_{2} \mathrm{O} \\ &2 \mathrm{ZnSO}_{4}+2 \mathrm{H}_{2} \mathrm{O} \stackrel{80 \%}{\longrightarrow} 2 \mathrm{Zn}+2 \mathrm{H}_{2} \mathrm{SO}_{4}+\mathrm{O}_{2} \end{aligned} $$ The number of moles of \(\mathrm{ZnS}\) required for producing 2 moles of \(\mathrm{Zn}\) will be : (a) \(3.125\) (b) 2 (c) \(2.125\) (d) 4

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A sample of peanut oil weighing \(2 \mathrm{~g}\) is added to \(25 \mathrm{~mL}\) of \(0.40 \mathrm{M} \mathrm{KOH}\). After saponification is complete, \(8.5 \mathrm{~mL}\) of \(0.28 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) is needed to neutralize excess of \(\mathrm{KOH}\). The saponification number of peanut oil is : (saponification number is defined as the milligrams of KOH consumed by \(1 \mathrm{~g}\) of oil) (a) \(146.72\) (b) \(223.44\) (c) \(98.9\) (d) None of these

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