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Chapter 11: Problem 88

What is the molar mass of a gas if \(1.15 \mathrm{~g}\) of the gas has a volume of \(225 \mathrm{~mL}\) at \(0^{\circ} \mathrm{C}\) and \(1.00 \mathrm{~atm}(\mathrm{STP}) ?(11.8)\)

Short Answer

Expert verified
The molar mass of the gas is approximately 113.86 g/mol.

Step by step solution

01

- Understand STP Conditions

Standard Temperature and Pressure (STP) conditions are defined as a temperature of 0掳C (273.15 K) and a pressure of 1 atm.
02

- Use the Ideal Gas Law

The Ideal Gas Law is written as \[ PV = nRT \] where \( P \) is pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the universal gas constant (8.314 J/(mol K) or 0.0821 atm L/(mol K)), and \( T \) is temperature in kelvin.
03

- Convert Volume to Liters

The given volume is 225 mL (milliliters). Convert it to liters by dividing by 1000: \[ V = 225 \text{ mL} \times \frac{1 \text{ L}}{1000 \text{ mL}} = 0.225 \text{ L} \]
04

- Convert Celsius to Kelvin

Convert the temperature from Celsius to Kelvin for use in the Ideal Gas Law: \[ T = 0^{\text{o}} \text{C} + 273.15 = 273.15 \text{ K} \]
05

- Solve for Number of Moles (n)

Rearrange the Ideal Gas Law to solve for the number of moles \( n \): \[ n = \frac{PV}{RT} \]Substitute the known values: \[ n = \frac{(1.00 \text{ atm})(0.225 \text{ L})}{(0.0821 \text{ atm L/mol K})(273.15 \text{ K})} \]Calculate \( n \):\[ n \approx 0.0101 \text{ moles} \]
06

- Calculate Molar Mass (M)

The molar mass \( M \) can be found using the formula: \[ M = \frac{ \text{mass of gas} }{ \text{number of moles} } = \frac{1.15 \text{ g}}{0.0101 \text{ moles}} \]Calculate the molar mass: \[ M \approx 113.86 \text{ g/mol} \]

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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 like a roadmap for understanding how gases behave under different conditions. It's written as: \[ PV = nRT \].
Each letter stands for something important:
  • P represents pressure
  • V is volume
  • n is the number of moles, which measures the amount of substance.
  • R is the universal gas constant (8.314 J/(mol路K) or 0.0821 atm路L/(mol路K)), and
  • T stands for temperature in Kelvin.
The equation tells us that pressure and volume are directly proportional to the number of moles and temperature. If you know the pressure, volume, and temperature of a gas, you can find out how many moles of the gas you have by rearranging the equation to: \[ n = \frac{PV}{RT} \]. It鈥檚 a neat way to predict how a gas will behave!
STP conditions
STP stands for Standard Temperature and Pressure. These conditions offer a reference point for scientists studying gases. They are defined as:
  • Temperature: 0掳C (273.15 K)
  • Pressure: 1 atm
Using STP conditions makes it easier to compare different gases because you鈥檙e always working with the same baseline. When the question tells us that a gas is at STP, it鈥檚 giving us a lot of information in a few words! Fourth, it's almost like working with standardized settings in a lab, so your results can be easily used and understood by others.
Unit conversions
Getting your units right is crucial in calculations.
Here鈥檚 why unit conversions matter:
  • Gases are often measured in different units, like liters and milliliters for volume or degrees Celsius and Kelvin for temperature.
  • For the Ideal Gas Law, we need specific units (liters and Kelvin) to make the equation work correctly.
For example, if volume is given in milliliters, we convert it to liters by dividing by 1000:
  • \[ V = 225 \text{ mL} \times \frac{1 \text{ L}}{1000 \text{ mL}} = 0.225 \text{ L} \]
Similarly, to convert Celsius to Kelvin, we use:
  • \[ T = 0^{\text{o}} \text{C} + 273.15 = 273.15 \text{ K} \]
Once everything is in the right units, the calculations flow smoothly and correctly.
Moles calculation
Calculating the number of moles is a key step in chemistry problems. Moles tell us how much of a substance we have. To find out the number of moles of gas using the Ideal Gas Law, we rearrange the equation:
\[ n = \frac{PV}{RT} \].
By plugging in the values from our problem:
  • Pressure (P): 1.00 atm
  • Volume (V): 0.225 L
  • Gas constant (R): 0.0821 atm路L/(mol路K)
  • Temperature (T): 273.15 K
We calculate:
\[ n = \frac{(1.00 \text{ atm})(0.225 \text{ L})}{(0.0821 \text{ atm L/mol K})(273.15 \text{ K})} \approx 0.0101 \text{ moles} \]. Once we have the moles, we can find other important values, like the molar mass:
\[ M = \frac{ \text{mass of gas} }{ \text{number of moles} } = \frac{1.15 \text{ g}}{0.0101 \text{ moles}} \approx 113.86 \text{ g/mol} \].

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Most popular questions from this chapter

When heated to \(350{ }^{\circ} \mathrm{C}\) at \(0.950\) atm, ammonium nitrate decomposes to produce nitrogen, water, and oxygen gases. $$ 2 \mathrm{NH}_{4} \mathrm{NO}_{3}(s) \stackrel{\Delta}{\longrightarrow} 2 \mathrm{~N}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{O}_{2}(g) $$ a. How many liters of water vapor are produced when \(25.8 \mathrm{~g}\) of \(\mathrm{NH}_{4} \mathrm{NO}_{3}\) decomposes? b. How many grams of \(\mathrm{NH}_{4} \mathrm{NO}_{3}\) are needed to produce \(10.0 \mathrm{~L}\) of oxygen?

On a climb up Mount Whitney, the atmospheric pressure drops to \(467 \mathrm{mmHg}\). What is the pressure in terms of the following units? a. atm b. torr c. in. \(\mathrm{Hg}\) d. \(\mathrm{Pa}\)

A gas mixture with a total pressure of 2400 torr is used by a scuba diver. If the mixture contains \(2.0 \mathrm{~mol}\) of helium and \(6.0\) mol of oxygen, what is the partial pressure, in torr, of each gas in the sample? ( \(11.10\) )

Glucose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\), is metabolized in living systems. How many grams of water can be produced from the reaction of \(18.0 \mathrm{~g}\) of glucose and \(7.50 \mathrm{~L}\) of \(\mathrm{O}_{2}\) at \(1.00 \mathrm{~atm}\) and \(37^{\circ} \mathrm{C}\) ? \((9.2,9.3,11.8,11.9)\) $$ \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g) \longrightarrow 6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) $$

Two flasks of equal volume and at the same temperature contain different gases. One flask contains \(5.0 \mathrm{~g}\) of \(\mathrm{O}_{2}\), and the other flask contains \(5.0 \mathrm{~g}\) of \(\mathrm{H}_{2} .\) Is each of the following statements true or false? Explain. (11.1) a. Both flasks contain the same number of molecules. b. The pressures in the flasks are the same.
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