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Chapter 1: Problem 20

A gas sample with a mass of 10 grams occupies 5.0 liters and exerts a pressure of 2.0 atm at a temperature of \(26^{\circ} \mathrm{C} .\) Which of the following expressions is equal to the molecular mass of the gas? The gas constant, \(R,\) is \(0.08(\mathrm{L} \times \mathrm{atm}) / \mathrm{mol} \times \mathrm{K}\) ). (A) \((0.08)(299) \mathrm{g} / \mathrm{mol}\) (B) \(\frac{(299)(0.50)}{(2.0)(0.08)} \mathrm{g} / \mathrm{mol}\) (C) \(\frac{299}{0.08} \mathrm{g} / \mathrm{mol}\) (D) \((2.0)(0.08) \mathrm{g} / \mathrm{mol}\)

Short Answer

Expert verified
None of the options is correct. The molar mass of the gas, calculated using the provided data and formula, is approximately 30 g/mol, which none of the choices matches.

Step by step solution

01

Convert Celsius to Kelvin

Temperature must be represented in Kelvin for the Ideal Gas Law. The conversion formula is \(K = ^{\circ}C + 273\). Therefore, convert the given Celsius temperature to Kelvin: \(K = 26^{\circ}C + 273 = 299K\).
02

Apply the Ideal Gas Law

Rearrange the Ideal Gas Law to solve for the number of moles, so that it becomes \(n = \frac{PV}{RT}\). Substituting the given values into the equation, we get \(n = \frac{(2.0 atm)(5.0 L)}{(0.08 L \cdot atm/mol \cdot K)(299 K)} = 0.333 mol\).
03

Calculate the Molar Mass

Use the formula for molar mass, \(M = \frac{m}{n}\), to find it. Substituting the calculated number of moles and given mass of the gas into the formula we obtain \(M = \frac{10 g}{0.333 mol} = 30 g/mol\).
04

Match the Result to the Choices

The goal is to find which of the choices matches with the calculated molar mass when calculated. None of them matches, they represent different calculations. However, if we look at choice (B), its formula looks like the formula used to solve for the number of moles (not molar mass), \(n = \frac{PV}{RT}\). If we rewrite it as molar mass \(M = \frac{m}{n}\), it becomes \(M = \frac{(2.0 atm)(0.08 L \cdot atm/mol \cdot K)}{(299 K)(0.5 mol)}\), which is still not matching the correct calculation for molar mass. So none of the options are correct for the molar mass of the gas considering the values provided in the question.

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Key Concepts

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

Molecular Mass Calculation
Molecular mass calculation is crucial in chemistry, especially when working with gases and the Ideal Gas Law. It allows scientists to determine the identity of a gas or confirm its purity. In simple terms, the molecular mass of a substance is the mass of a molecule of the substance in atomic mass units.

To calculate the molecular mass, we usually convert it into a measurable form like molar mass, which is the mass of a mole of molecules in grams per mole (\text{g/mol}). The molar mass formula is given by: \[ M = \frac{m}{n} \] where \( M \) is the molar mass, \( m \) is the mass in grams, and \( n \) is the number of moles.

In the solved exercise, we used this formula to find the molar mass by dividing the total mass of gas by the number of moles: \[ M = \frac{10 \text{ g}}{0.333 \text{ mol}} = 30 \text{ g/mol} \] This calculation can be helpful in identifying the gas based on its molar mass. Always ensure to have your mass and moles accurately measured to get the correct molecular mass.
Temperature Conversion to Kelvin
Temperature conversion is essential for calculations in the Ideal Gas Law since it requires temperature in Kelvin. Kelvin is the absolute temperature scale widely used in scientific calculations. Unlike Celsius, Kelvin does not go below zero, helping avoid negative temperatures in equations.

To convert Celsius to Kelvin, you use the formula: \[ K = ^{\circ}C + 273 \] In the original exercise, we had a temperature of \( 26^{\circ}C \). Converting this to Kelvin gives: \[ K = 26 + 273 = 299 \text{ K} \]

By converting to Kelvin, we ensure a smooth and error-free application of the Ideal Gas Law, which simplifies and standardizes calculations involving temperatures. Always double-check your temperature unit, as this conversion step is fundamental in gas law equations.
Molar Mass Formula
The molar mass formula is pivotal when analyzing gases and using the Ideal Gas Law. It effectively links the mass of a substance to its mole content, providing comprehensive information about the gas itself.

The molar mass, represented in \text{g/mol}, is expressed by the formula: \[ M = \frac{m}{n} \] where \( M \) is the molar mass, \( m \) is the mass of the gas in grams, and \( n \) is the number of moles.

Using the Ideal Gas Law, \( PV = nRT \), we rearrange it to find the number of moles: \[ n = \frac{PV}{RT} \] Substitute this back into the molar mass equation allows one to find the molar mass when provided with measurable quantities like pressure, volume, and temperature of the gas.

Calculating molar mass guides us in determining unknown gas samples and evaluating chemical reactions, making it an indispensable tool in both academic and industrial chemistry settings. Always ensure units are consistent when performing these calculations, to get accurate and reliable results.

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Most popular questions from this chapter

Use the following information to answer questions 29-31. Pennies are made primarily of zinc, which is coated with a thin layer of copper through electroplating, using a setup like the one above. The solution in the beaker is a strong acid (which produces H' ions), and the cell is wired so that the copper electrode is the anode and zinc penny is the cathode. Use the following reduction potentials to answer questions \(29-31 .\) $$\begin{array}{|l|l|}\hline \text { Half-Reaction } & {\text { Standard Reduction Potential }} \\ \hline \mathrm{Cu}^{2++2 e^{-} \rightarrow \mathrm{Cu}(s)} & {+0.34 \mathrm{V}} \\ \hline 2 \mathrm{H}^{++2 e^{-} \rightarrow \mathrm{H}_{2}(g)} & {0.00 \mathrm{V}} \\ \hline \mathrm{Ni}^{2++2 e^{-} \rightarrow \mathrm{Ni}(s)} & {-0.25 \mathrm{V}} \\\ \hline \mathrm{Zn}^{2++2 e^{-} \rightarrow \mathrm{Zn}(s)} & {-0.76 \mathrm{V}} \\ \hline\end{array}$$ What is the required voltage to make this cell function? (A) 0.34 V (B) 0.42 V (C) 0.76 V (D) 1.10 V

Questions 54-56 refer to the following. GRAPH CAN'T COPY Between propane and ethene, which will likely have the higher boiling point and why? (A) Propane, because it has a greater molar mass (B) Propane, because it has a more polarizable electron cloud (C) Ethene, because of the double bond (D) Ethene, because it is smaller in size

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