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Chapter 10: Problem 54

The molar mass of a volatile substance was determined by the Dumas-bulb method described in Exercise 10.53 . The unknown vapor had a mass of \(2.55 \mathrm{~g} ;\) the volume of the bulb was \(500 \mathrm{~mL}\), pressure \(101.33 \mathrm{kPa}\), and temperature \(37^{\circ} \mathrm{C} .\) Calculate the molar mass of the unknown vapor.

Short Answer

Expert verified
The molar mass of the unknown vapor is approximately 129.44 g/mol.

Step by step solution

01

Convert Units Where Necessary

First, convert the volume from mL to L, and the temperature from Celsius to Kelvin. Since 1 L equals 1000 mL, the volume is \( 500 \, \text{mL} = 0.500 \, \text{L} \). The temperature in Kelvin is \( 37^{\circ} \text{C} + 273.15 = 310.15 \text{K} \).
02

Use the Ideal Gas Law

Apply the ideal gas law equation, \( PV = nRT \), where \( P \) is the pressure in pascals (Pa), \( V \) is volume in liters (L), \( n \) is the amount of substance in moles, \( R \) is the ideal gas constant \(8.314\text{ J/mol K} \), and \( T \) is the temperature in Kelvin. First, convert pressure from kPa to Pa: \( 1 \text{ kPa} = 1000 \text{ Pa} \), so \( 101.33 \text{ kPa} = 101330 \text{ Pa} \).
03

Solve for Number of Moles

Rearrange the equation to solve for \( n \): \( n = \frac{PV}{RT} \). Substituting the values, we have \( n = \frac{101330 \, \text{Pa} \times 0.500 \, \text{L}}{8.314 \, \text{J/mol K} \times 310.15 \, \text{K}} \approx 0.0197 \, \text{mol} \).
04

Calculate Molar Mass

The molar mass \( M \) is calculated by dividing the mass of the gas by the number of moles: \( M = \frac{\text{mass}}{n} \). Given the mass is \( 2.55 \, \text{g} \), it follows that \( M = \frac{2.55 \, \text{g}}{0.0197 \, \text{mol}} \approx 129.44 \, \text{g/mol} \).

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Key Concepts

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

Ideal Gas Law
The Ideal Gas Law is a vital formula in chemistry that combines several gas laws into one. It is given by the equation \( PV = nRT \), where:
  • \( P \) represents pressure, measured in pascals (Pa).
  • \( V \) stands for volume, measured in liters (L).
  • \( n \) is the number of moles of gas.
  • \( R \) is the ideal gas constant, which has a value of \( 8.314 \, \text{J/mol K} \).
  • \( T \) is the temperature in Kelvin.
The law allows you to relate these variables for a hypothetical ideal gas. It is invaluable for calculating properties of gases, like determining the number of moles in the given problem. First, make sure all units are in the correct form, such as converting pressure from kPa to Pa using 1 kPa = 1000 Pa. The equation helps in calculating moles, which is a crucial step in finding the molar mass.
Temperature Conversion
Temperature is often given in Celsius for everyday usage, but in scientific calculations, it's usually needed in Kelvin. Kelvin is the SI unit for temperature and is essential in gas law calculations since it avoids negative temperatures. To convert Celsius to Kelvin, simply add 273.15.
  • For example, a temperature of \( 37^{\circ} \text{C} \) is converted to Kelvin by adding 273.15, resulting in \( 310.15 \text{K} \).
Always ensure temperature is in Kelvin when using the Ideal Gas Law, since the gas constant \( R \) is defined using Kelvin. This helps maintain consistency across the units and ensures accurate calculations.
Dumas-bulb Method
The Dumas-bulb method is a classical technique used to determine the molar mass of a volatile liquid. This method involves vaporizing the liquid in a bulb and applying the Ideal Gas Law. Here鈥檚 how it works:
  • The substance is heated until it converts to vapor and fills the bulb.
  • Once the bulb is filled, it is cooled, and the remaining vapor is measured in terms of volume, pressure, and temperature.
  • The weight of the condensed liquid provides the mass of the gas.
With these measurements, you can calculate the number of moles using \( PV = nRT \), where all conditions are met. Finally, divide the mass by the moles to get molar mass. This method offers a practical application of the Ideal Gas Law and is primarily used for determining the properties of unknown substances.
Unit Conversion
Unit conversion is a fundamental step in solving problems related to gas laws. Proper conversion ensures consistency across the equations and prevents errors. In context, different units might be used in the given data, such as:
  • Volume: Convert from mL to L by dividing by 1000.
  • Pressure: Convert from kPa to Pa by multiplying by 1000, ensuring that pressure is in the correct SI unit for use in gas law equations.
  • Mass: Ensure that it is in grams for compatibility with other units when calculating molar mass.
Being meticulous about using the right units for each part of the calculation is crucial for accuracy. Unit conversion allows seamless integration of given data into scientific equations. This step-by-step adjustment is a foundational skill in mastering chemistry and physics problems.

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Most popular questions from this chapter

Hurricane Wilma of 2005 is the most intense hurricane on record in the Atlantic basin, with a low-pressure reading of 882 mbar (millibars). Convert this reading into (a) atmospheres, \((\mathbf{b})\) torr, and \((\mathbf{c})\) inches of \(\mathrm{Hg}\).

Consider the following gases, all at STP: Ne, \(\mathrm{SF}_{6}, \mathrm{~N}_{2}, \mathrm{CH}_{4}\). (a) Which gas is most likely to depart from the assumption of the kinetic- molecular theory that says there are no attractive or repulsive forces between molecules? (b) Which one is closest to an ideal gas in its behavior? (c) Which one has the highest root-mean-square molecular speed at a given temperature? (d) Which one has the highest total molecular volume relative to the space occupied by the gas? (e) Which has the highest average kinetic- molecular energy? (f) Which one would effuse more rapidly than \(\mathrm{N}_{2} ?\) (g) Which one would have the largest van der Waals \(b\) parameter?

Does the effect of intermolecular attraction on the properties of a gas become more significant or less significant if (a) the gas is compressed to a smaller volume at constant temperature; (b) the temperature of the gas is increased at constant volume?

An 8.40 -g sample of argon and an unknown mass of \(\mathrm{H}_{2}\) are mixed in a flask at room temperature. The partial pressure of the argon is \(44.0 \mathrm{kPa}\), and that of the hydrogen is \(57.33 \mathrm{kPa}\). What is the mass of the hydrogen?

Propane, \(\mathrm{C}_{3} \mathrm{H}_{8}\), liquefies under modest pressure, allowing a large amount to be stored in a container. (a) Calculate the number of moles of propane gas in a 20 -L container at \(709.3 \mathrm{kPa}\) and \(25^{\circ} \mathrm{C} .\) (b) Calculate the number of moles of liquid propane that can be stored in the same volume if the density of the liquid is \(0.590 \mathrm{~g} / \mathrm{mL}\). (c) Calculate the ratio of the number of moles of liquid to moles of gas. Discuss this ratio in light of the kinetic-molecular theory of gases.
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