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Chapter 6: Problem 42

Monochloroethylene is used to make polyvinylchloride (PVC). It has a density of \(2.56 \mathrm{g} / \mathrm{L}\) at \(22.8^{\circ} \mathrm{C}\) and 756 mmHg. What is the molar mass of monochloroethylene? What is the molar volume under these conditions?

Short Answer

Expert verified
The molar mass of monochloroethylene is 0.000114 kg/mol, and its molar volume under the given conditions is 22.8 L/mol

Step by step solution

01

Finding the molar mass

Firstly, it is important to understand that the molar mass of a gas can be calculated by dividing its density by its molar volume. Since the molar volume of a gas at tandard conditions (which is approximately close to the conditions stated in this problem: 22.8 degrees Celsius and 756 mmHg) is \(22.4 L/mol\), the molar mass (M) of monochloroethylene can be calculated as follows: \(M = 2.56 g/L ÷ 22.4 L/mol = 0.114 g/mol\)
02

Convert g/mol to Kg/mol

Scientifically, the molar mass should be in Kg/mol, so the calculated molar mass needs to be converted from g/mol to kg/mol, by dividing it by 1000. Therefore, \(M = 0.114 g/mol ÷ 1000 = 0.000114 kg/mol\)
03

Calculate the molar volume

Secondly, you have to calculate the molar volume of the gas under the given conditions. To achieve this, use the ideal gas law formula: \(P×V=n×R×T\), where P is the pressure, V is the volume, n is the number of moles, R is the gas constant and T is the temperature in Kelvin. We know n/molar mass = density, and R = 0.0821 L·atm/(K·mol). We convert the pressure from mmHg to atm and temperature from Celsius to Kelvin, and then solve for volume, V. Hence, the molar volume (V) of the gas can be found as follows: \(V= ( nR_{t}T_{1} ) / P_{2} = (density × R × (22.8 + 273.15)) / (756 ÷ 760) = 22.8 L/mol\)

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Most popular questions from this chapter

In the reaction of \(\mathrm{CO}_{2}(\mathrm{g})\) and solid sodium peroxide \(\left(\mathrm{Na}_{2} \mathrm{O}_{2}\right),\) solid sodium carbonate \(\left(\mathrm{Na}_{2} \mathrm{CO}_{3}\right)\) and oxy- gen gas are formed. This reaction is used in submarines and space vehicles to remove expired \(\mathrm{CO}_{2}(\mathrm{g})\) and to generate some of the \(\mathrm{O}_{2}(\mathrm{g})\) required for breathing. Assume that the volume of gases exchanged in the lungs equals \(4.0 \mathrm{L} / \mathrm{min},\) the \(\mathrm{CO}_{2}\) content of expired air is \(3.8 \% \mathrm{CO}_{2}\) by volume, and the gases are at \(25^{\circ} \mathrm{C}\) and \(735 \mathrm{mmHg}\). If the \(\mathrm{CO}_{2}(\mathrm{g})\) and \(\mathrm{O}_{2}(\mathrm{g})\) in the above reaction are measured at the same temperature and pressure, (a) how many milliliters of \(\mathrm{O}_{2}(\mathrm{g})\) are produced per minute and \((\mathrm{b})\) at what rate is the \(\mathrm{Na}_{2} \mathrm{O}_{2}(\mathrm{s})\) consumed, in grams per hour?

Appendix E describes a useful study aid known as concept mapping. Using the method presented in Appendix \(\mathrm{E}\), construct a concept map illustrating the different concepts to show the relationships among all the gas laws described in this chapter.

A sample of \(\mathrm{N}_{2}(\mathrm{g})\) effuses through a tiny hole in \(38 \mathrm{s}\) What must be the molar mass of a gas that requires \(64 \mathrm{s}\) to effuse under identical conditions?

What are the ratios of the diffusion rates for the pairs of gases (a) \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2} ;\) (b) \(\mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{D}_{2} \mathrm{O}\) \(\left(\mathrm{D}=\text { deuterium, i.e., }_{1}^{2} \mathrm{H}\right) ;\) (c) \(^{14} \mathrm{CO}_{2}\) and \(^{12} \mathrm{CO}_{2}\) (d) \(^{235} \mathrm{UF}_{6}\) and \(^{238} \mathrm{UF}_{6} ?\)

Calculate \(u_{\mathrm{rms}},\) in meters per second, for \(\mathrm{Cl}_{2}(\mathrm{g})\) molecules at \(30^{\circ} \mathrm{C}\)
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