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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

Directions: Questions 1-3 are long free-response questions that require about 23 minutes each to answer and are worth 10 points each. Write your response in the space provided following each question. Examples and equations may be included in your responses where appropriate. For calculations, clearly show the method used and the steps involved in arriving at your answers. You must show your work to receive credit for your answer. Pay attention to significant figures. A student is tasked with determining the identity of an unknown carbonate compound with a mass of 1.89 g. The compound is first placed in water, where it dissolves completely. The \(K_{s p}\) value for several carbonate-containing compounds are given below. $$\begin{array}{|c|c|}\hline \text { Compound } & {K_{s p}} \\ \hline \text { Lithium carbonate } & {8.15 \times 10^{-4}} \\ \hline \text { Nickel (II) carbonate } & {1.42 \times 10^{-7}} \\ \hline \text { Strontium carbonate } & {5.60 \times 10^{-10}} \\ \hline\end{array}$$ (a) In order to precipitate the maximum amount of the carbonate ions from solution, which of the following should be added to the carbonate solution: \(\operatorname{LiNO}_{3}, \mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2},\) or \(\mathrm{Sr}\left(\mathrm{NO}_{3}\right)_{2} ?\) Justify your answer. (b) For the carbonate compound that contains the cation chosen in part (a), determine the concentration of each ion of that compound in solution at equilibrium. (c) When mixing the solution, should the student ensure the carbonate solution or the nitrate solution is in excess? Justify your answer. (d) After titrating sufficient solution to precipitate out all of the carbonate ions, the student filters the solution before placing it in a crucible and heating it to drive off the water. After several heatings, the final mass of the precipitate remains constant and is determined to be 2.02 g. (i) Determine the number of moles of precipitate. (ii) Determine the mass of carbonate present in the precipitate. (e) Determine the percent, by mass, of carbonate in the original sample. (f) Is the original compound most likely lithium carbonate, sodium carbonate, or potassium carbonate? Justify your answer.

Directions: Questions 4-7 are short free-response questions that require about 9 minutes each to answer and are worth 4 points each. Write your response in the space provided following each question. Examples and equations may be included in your responses where appropriate. For calculations, clearly show the method used and the steps involved in arriving at your answers. You must show your work to receive credit for your answer. Pay attention to significant figures. A stock solution of \(2.0 \mathrm{M} \mathrm{MgCl}_{2}\) is dissolved in water. (a) (i) In the beaker below, draw a particulate diagram that represents \(\mathrm{MgCl}_{2}\) dissolved in water. The approximate sizes of each atom/ion are provided for you. Your diagram should include at least four water molecules, which should be correctly oriented compared to the ions dissolved in solution. (DIAGRAM CANT COPY) (ii) Why are the chloride ions from (a)(i) larger than the magnesium (b) (i) A student wishes to make up 500 \(\mathrm{mL}\) of 0.50 \(M \mathrm{MgCl}_{2}\) for an experiment. Explain the best method of doing so utilizing a graduated cylinder and a volumetric flask. Assume \(\mathrm{MgCl}_{2}\) is fully soluble. (ii) What are the concentrations of the \(\mathrm{Mg}^{2+}\) and \(\mathrm{Cl}^{-}\) ions in the new solution?

The first ionization energy for a neutral atom of chlorine is 1.25 \(\mathrm{MJ} / \mathrm{mol}\) and the first ionization energy for a neutral atom of argon is 1.52 \(\mathrm{MJ} / \mathrm{mol}\) How would the first ionization energy value for a neutral atom of potassium compare to those values? (A) It would be greater than both because potassium carries a greater nuclear charge then either chlorine or argon. (B) It would be greater than both because the size of a potassium atom is smaller than an atom of either chlorine or argon. (C) It would be less than both because there are more electrons in potassium, meaning they repel each other more effectively and less energy is needed to remove one. (D) It would be less than both because a valence electron of potassium is farther from the nucleus than one of either chlorine or argon.

Which of the following could be added to an aqueous solution of weak acid HF to increase the percent dissociation? (A) \(\operatorname{NaF}(s)\) (B) \(\mathrm{H}_{2} \mathrm{O}(l)\) (C) \(\mathrm{NaOH}(\mathrm{s})\) (D) \(\mathrm{NH}_{3}(a q)\)

$$\begin{array}{|c|c|}\hline \text { Time (Hours) } & {[\mathrm{A}] M} \\\ \hline 0 & {0.40} \\ \hline 1 & {0.20} \\ \hline 2 & {0.10} \\ \hline 3 & {0.05} \\ \hline\end{array}$$ Reactant A underwent a decomposition reaction. The concentration of A was measured periodically and recorded in the chart above. Based on the data in the chart, which of the following is the rate law for the reaction? (A) Rate \(=k[\mathrm{A}]\) (B) Rate \(=k[\mathrm{A}]^{2}\) (C) Rate \(=2 k[\mathrm{A}]\) (D) Rate \(=\frac{1}{2} k[\mathrm{A}]\)
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