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Chapter 8: Problem 56

The density (in \(\mathrm{g} /\) l) of an equimolar mixture of methane and ethane at 1 atm and \(0^{\circ} \mathrm{C}\) is (a) \(1.03\) (b) \(2.05\) (c) \(0.94\) (d) \(1.25\)

Short Answer

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g/L

Step by step solution

01

1. Determine the molar mass of each gas

To start, we need to know the molar mass of both methane (CH4) and ethane (C2H6). From the periodic table, the molar mass of carbon is 12, and hydrogen is 1. Molar mass of methane (CH4) = 12 + (4 × 1) = 16 g/mol Molar mass of ethane (C2H6) = (2 × 12) + (6 × 1) = 24 + 6 = 30 g/mol
02

2. Find the molar mass of the equimolar mixture

Since the mixture contains equal moles of methane and ethane, the average molar mass of the mixture can be found by taking the arithmetic mean of the molar masses of both gases. Mixture molar mass = (Molar mass of CH4 + Molar mass of C2H6) / 2 Mixture molar mass = (16 g/mol + 30 g/mol) / 2 = 23 g/mol
03

3. Use the ideal gas equation to find the density of the mixture

The ideal gas equation is given by: PV = nRT where P is the pressure, V is the volume, n is the number of moles, R is the universal gas constant, and T is the temperature. To find the density (mass/volume), we can rearrange this equation: Density = mass/volume = (n × Mixture molar mass) / V Since PV = nRT, we can substitute for n: Density = (P × Mixture molar mass) / (R × T) Given: P = 1 atm, T = \(0^{\circ} \mathrm{C}\) = 273.15 K, and R = 0.0821 L × atm / (mol × K) Density = (1 atm × 23 g/mol) / (0.0821 L × atm / (mol × K) × 273.15 K)
04

4. Calculate the density and match with the given options

Now, we can calculate the density of the equimolar mixture: Density = (23.0 g/mol) / (0.0821 L × atm / (mol × K) × 273.15 K) = 1.03 g/L Comparing the calculated density with the given options, we find that the correct answer is: (a) \(1.03\)

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Most popular questions from this chapter

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