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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

$$\begin{array}{|c|c|}\hline \text { Time (Hours) } & {[\mathrm{A}] M} \\\ \hline 0 & {0.40} \\ \hline 1 & {0.20} \\ \hline 2 & {0.10} \\ \hline 3 & {0.05} \\ \hline\end{array}$$ 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 \(=\frac{1}{2} k[\mathrm{A}]\)

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. The unbalanced reaction between potassium permanganate and acidified iron (II) sulfate is a redox reaction that proceeds as follows: $$\mathrm{H}^{+}(a q)+\mathrm{Fe}^{2+}(a q)+\mathrm{MnO}_{4}^{-(a q)} \rightarrow \mathrm{Mn}^{2+}(a q)+\mathrm{Fe}^{3+}(a q)+\mathrm{H}_{2} \mathrm{O}(l)$$ (a) Provide the equations for both half-reactions that occur below: (i) Oxidation half-reaction (ii) Reduction half-reaction (b) What is the balanced net ionic equation? A solution of 0.150 M potassium permanganate is placed in a buret before being titrated into a flask containing 50.00 mL of iron (II) sulfate solution of unknown concentration. The following data describes the colors of the various ions in solution: $$\begin{array}{|c|c|}\hline \text { Ion } & {\text { Color in solution }} \\\ \hline \mathrm{H^{+ }} & {\text { Colorless }} \\ \hline \mathrm{Fe}^{2+} & {\text { Pale Green }} \\ \hline \mathrm{MnO}_{4^{-}} & {\text {Dark Purple }} \\ \hline \mathrm{Mn}^{2+} & {\text { Colorless }} \\ \hline \mathrm{Fe}^{3+} & {\text { Yellow }} \\ \hline \mathrm{K}^{+} & {\text {Colorless }} \\ \hline \mathrm{SO}_{4}^{2-} & {\text { Colorless }} \\\ \hline\end{array}$$ (c) Describe the color of the solution in the flask at the following points: (i) Before titration begins (ii) During titration prior to the endpoint (iii) At the endpoint of the titration (d) (i) If 15.55 mL of permanganate are added to reach the endpoint, what is the initial concentration of the iron (II) sulfate? (ii) The actual concentration of the \(\mathrm{FeSO}_{4}\), is 0.250 \(M\) . Calculate the percent error. (e) Could the following errors have led to the experimental result deviating in the direction that it did? You must justify your answers quantitatively. (i) 55.0 \(\mathrm{mL}\) of \(\mathrm{FeSO}_{4}\) was added to the flask prior to titration instead of 50.0 mL . (ii) The concentration of the potassium permanganate was actually 0.160 \(M\) instead of 0.150 \(M\) .

Which compound, \(\mathrm{CaCl}_{2}\) or \(\mathrm{CaO}\) , would you expect to have a high melting point? Why? (A) \(\mathrm{CaCl}_{2}\) because there are more ions per lattice unit (B) \(\mathrm{CaCl}_{2}\) because a chlorine ion is smaller than an oxygen ion (C) Cao, because the charge of oxygen ion exceeds that of chlorine ion (D) CaO, because the common charges of calcium and oxygen ions are identical in magnitude

Which of the following pairs of elements is most likely to create an interstitial alloy? (A) Titanium and copper (B) Aluminum and lead (C) Silver and tin (D) Magnesium and calcium

Which of the following compounds would have the highest lattice energy? (A) LiF (B) MgCl_ (C) CaBr_ (D) \(\mathrm{C}_{2} \mathrm{H}_{6}\)
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