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Chapter 5: Problem 47

A 2.10-L vessel contains 4.65 g of a gas at 1.00 atm and \(27.0^{\circ} \mathrm{C}\). (a) Calculate the density of the gas in grams per liter. (b) What is the molar mass of the gas?

Short Answer

Expert verified
The density of the gas is 2.21 g/L and the molar mass of the gas is 54.7 g/mol.

Step by step solution

01

Convert Temperature from Celsius to Kelvin

Using the conversion \(K = 掳C + 273.15\), the provided temperature value of \(27.0 掳C\) becomes \(27.0 掳C + 273.15 = 300.15 K\).
02

Find Gas Density

The gas density formula is \(D = \frac{m}{V}\), where \(D\) is density, \(m\) is the mass of gas (4.65 g), and \(V\) is the volume of gas (2.10 L). So, \(D = \frac{4.65 g}{2.10 L} = 2.21 g/L\).
03

Determine Molar Mass using the Ideal Gas Law

The ideal gas law formula is \(PV = nRT\), where \(P\) is pressure (1.00 atm), \(V\) is volume (2.10 L), \(n\) is number of moles, \(R\) is gas constant (0.0821 \(L \cdot atm / (K \cdot mol\))), and \(T\) is temperature (300.15 K). To calculate the number of moles \(n\), it's rearranged to \(n = \frac{PV}{RT}\). Substituting, \(n = \frac{1.00 atm \cdot 2.10 L}{0.0821 L \cdot atm / (K \cdot mol) \cdot 300.15 K} = 0.085 mol\). The formula for molar mass is \(MM = \frac{m}{n}\), where \(m\) is the mass of gas (4.65 g) and \(n\) is number of moles (0.085 mol). Substituting the values, \(MM = \frac{4.65 g}{0.085 mol} = 54.7 g/mol\).

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Key Concepts

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

Density Calculation of a Gas
Calculating the density of a gas involves using a simple formula. The density \(D\) of a gas is determined by dividing its mass \(m\) by the volume \(V\) it occupies. In the ideal gas law problem given, you have a gas with a mass of 4.65 grams, contained in a 2.10-liter vessel. Applying the formula:
  • \(D = \frac{m}{V}\)
Substituting in the numbers:
  • \(D = \frac{4.65 \text{ g}}{2.10 \text{ L}}\)
  • \(D \approx 2.21 \text{ g/L}\)
This calculation gives you the density of the gas in grams per liter. It basically tells you how much mass of the gas is present in each liter of volume it occupies.
Molar Mass Determination
Determining the molar mass of a gas in this context uses the ideal gas law, which is a cornerstone equation in chemistry. The ideal gas law is \(PV = nRT\), where \(P\) is pressure, \(V\) is volume, \(n\) is the number of moles, \(R\) is the ideal gas constant, and \(T\) is the temperature in Kelvin. To find the number of moles of the gas, rearrange the equation to \(n = \frac{PV}{RT}\).
  • Given: \(P = 1.00 \text{ atm}\), \(V = 2.10 \text{ L}\), \(R = 0.0821 \text{ L} \cdot \text{atm} / (\text{K} \cdot \text{mol})\), and \(T = 300.15 \text{ K}\).
  • \(n = \frac{1.00 \text{ atm} \times 2.10 \text{ L}}{0.0821 \text{ L} \cdot \text{atm} / (\text{K} \cdot \text{mol}) \times 300.15 \text{ K}}\)
  • \(n \approx 0.085 \text{ mol}\)
With the number of moles determined, calculate the molar mass \(MM\) using \(MM = \frac{m}{n}\), where \(m\) is mass and \(n\) is moles.
  • \(MM = \frac{4.65 \text{ g}}{0.085 \text{ mol}}\)
  • \(MM \approx 54.7 \text{ g/mol}\)
This tells you the mass of one mole of the gas.
Temperature Conversion to Kelvin
Converting temperature to Kelvin is straightforward yet essential in calculations involving gases, as the Kelvin scale is used in gas equations to maintain proportional relationships. The conversion formula from Celsius to Kelvin is:
  • \(K = 掳C + 273.15\)
For the given problem, the initial temperature is 27.0掳C. Applying the conversion:
  • \(K = 27.0 \degree C + 273.15\)
  • \(K = 300.15\)
The Kelvin temperature is now suitable for use in the ideal gas law and other thermodynamic equations. This process ensures that all temperature measurements align with the absolute scale, which starts at absolute zero, making it ideal for scientific calculations.

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Most popular questions from this chapter

A gaseous hydrocarbon (containing C and H atoms) in a container of volume \(20.2 \mathrm{~L}\) at \(350 \mathrm{~K}\) and 6.63 atm reacts with an excess of oxygen to form \(205.1 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) and \(168.0 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O} .\) What is the molecular formula of the hydrocarbon?

A sample of nitrogen gas kept in a container of volume \(2.3 \mathrm{~L}\) and at a temperature of \(32^{\circ} \mathrm{C}\) exerts a pressure of 4.7 atm. Calculate the number of moles of gas present.

Two vessels are labeled A and B. Vessel A contains \(\mathrm{NH}_{3}\) gas at \(70^{\circ} \mathrm{C},\) and vessel \(\mathrm{B}\) contains \(\mathrm{Ne}\) gas at the same temperature. If the average kinetic energy of \(\mathrm{NH}_{3}\) is \(7.1 \times 10^{-21} \mathrm{~J} / \mathrm{molecule},\) calculate the rootmean- square speed of Ne atoms.

The gas laws are vitally important to scuba divers. The pressure exerted by \(33 \mathrm{ft}\) of seawater is equivalent to 1 atm pressure. (a) A diver ascends quickly to the surface of the water from a depth of \(36 \mathrm{ft}\) without exhaling gas from his lungs. By what factor will the volume of his lungs increase by the time he reaches the surface? Assume that the temperature is constant. (b) The partial pressure of oxygen in air is about 0.20 atm. (Air is 20 percent oxygen by volume.) In deep-sea diving, the composition of air the diver breathes must be changed to maintain this partial pressure. What must the oxygen content (in percent by volume) be when the total pressure exerted on the diver is 4.0 atm? (At constant temperature and pressure, the volume of a gas is directly proportional to the number of moles of gases.) (Hint: See the Chemistry in Action essay "Scuba Diving and the Gas Laws" in Section \(5.6 . ?\)

Which of the following statements is correct? (a) Heat is produced by the collision of gas molecules against one another. (b) When a gas is heated, the molecules collide with one another more often.
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