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Chapter 10: Problem 51

(a) Calculate the density of \(\mathrm{NO}_{2}\) gas at \(0.970\) atm and \(35^{\circ} \mathrm{C}\). (b) Calculate the molar mass of a gas if \(2.50 \mathrm{~g}\) occupies \(0.875 \mathrm{~L}\) at 685 torr and \(35^{\circ} \mathrm{C}\).

Short Answer

Expert verified
The density of \(\mathrm{NO}_{2}\) gas at \(0.970\) atm and \(35^{\circ} \mathrm{C}\) is \(1.80\,\text{g/L}\). The molar mass of the gas in the second problem is \(72.05\,\text{g/mol}\).

Step by step solution

01

Convert the temperature to Kelvin

To convert the temperature from Celsius to Kelvin, use the formula: \(T(K) = T(^\circ C) + 273.15\) So, for \(35^\circ C\), \(T(K) = 35 + 273.15 = 308.15\,\text{K}\)
02

Calculate the molar mass of \(\mathrm{NO}_{2}\)

To find the molar mass of \(\mathrm{NO}_{2}\), find the molar mass of each element in the compound and add them up: Molar mass of N: \(14.01\,\text{g/mol}\) Molar mass of O: \(16.00\,\text{g/mol}\) Molar mass of \(\mathrm{NO}_{2}\): \(14.01 + 2 \times 16.00 = 46.01\,\text{g/mol}\)
03

Calculate the density

Now, we will use the Ideal Gas Law and rearrange it to solve for the density, where \(density = \frac{Mass}{Volume}\), as well as \(n = \frac{Mass}{Molar~mass}\). Density = \(\frac{molar~mass \times P}{R \times T}\) Here, P is the pressure, R is the gas constant (0.08206 \(L\,atm/(mol\,K)\)), and T is the temperature in Kelvin. Density of \(\mathrm{NO}_{2}\) = \(\frac{46.01~g/mol \times 0.970~atm}{0.08206~\frac{L\,atm}{mol\,K} \times 308.15\,\text{K}} = 1.80\,\text{g/L}\) The density of \(\mathrm{NO}_{2}\) gas at \(0.970\) atm and \(35^{\circ} \mathrm{C}\) is \(1.80\,\text{g/L}\). ##Problem (b)##
04

Convert the pressure to atm and the temperature to Kelvin

To convert the pressure from torr to atm, use the formula: \(P(atm) = P(torr) \times \frac{1\,\text{atm}}{760\,\text{torr}}\) So, for 685 torr, \(P(atm) = 685 \times \frac{1}{760} = 0.9013\,\text{atm}\) To convert the temperature from Celsius to Kelvin, use the same formula as in problem (a): \(T(K) = 35 + 273.15 = 308.15\,\text{K}\)
05

Calculate the number of moles using the Ideal Gas Law

Rearrange the Ideal Gas Law to solve for the number of moles (n): \(n = \frac{PV}{RT}\) \(n = \frac{0.9013\,\text{atm} \times 0.875\,\text{L}}{0.08206\,(\frac{L\,atm}{mol\,K}) \times 308.15\,\text{K}} = 0.03470\,\text{mol}\)
06

Calculate the molar mass

Finally, use the formula \(Molar~mass = \frac{Mass}{n}\) to calculate the molar mass of the gas: Molar mass = \(\frac{2.50\,\text{g}}{0.03470\,\text{mol}} = 72.05\,\text{g/mol}\) The molar mass of the gas is \(72.05\,\text{g/mol}\).

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Key Concepts

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

Ideal Gas Law
Understanding the Ideal Gas Law is crucial for solving a wide range of problems involving gases. This law helps us relate the pressure (P), volume (V), temperature (T), and number of moles (n) of an ideal gas through the equation:
\[ PV = nRT \]
where R is the universal gas constant. The value of R depends on the units used for pressure, volume, and temperature. For instance, when pressure is in atmospheres (atm) and volume in liters (L), R is approximately 0.08206 L鈰卆tm/(mol鈰匥). Before using this law, make sure your temperature is in Kelvin (K), which requires conversion from Celsius (掳C) using the formula \( T(K) = T(掳C) + 273.15 \).
Temperature Conversion
To solve gas-related problems, it's essential to convert the temperature to Kelvin as the Ideal Gas Law requires the absolute temperature scale. The conversion is simple: Take the Celsius temperature and add 273.15 to it.
\[ T(K) = T(掳C) + 273.15 \]
For the problem above, the conversion from 35掳C yields \( T(K) = 35 + 273.15 = 308.15\text{K} \). Remember, Kelvin is the SI unit for temperature, and the Kelvin scale starts at absolute zero, the theoretically coldest possible temperature.
Pressure Conversion
In gas calculations, pressure may be provided in units like torr, but the Ideal Gas Law typically uses atmospheres (atm). Thus, converting pressure to the correct units is essential. To convert torr to atm, divide the torr value by 760, because 1 atm is equal to 760 torr.
\[ P(atm) = P(torr) \times \frac{1\text{atm}}{760\text{torr}} \]
In our exercise, converting 685 torr to atm gives:
\[ P(atm) = 685 \times \frac{1}{760} = 0.9013\text{atm} \].
Molar Mass Calculation
The molar mass of a substance is the mass of one mole of that substance, typically expressed in grams per mole (g/mol). It is a vital constant in gas calculations as it relates the mass of the gas to the number of moles. To calculate the molar mass, take the mass of the sample and divide it by the number of moles
\[ \text{Molar mass} = \frac{\text{Mass}}{\text{n}} \]
For the given problem, where 2.50 g of gas occupies 0.875 L at 685 torr and 35掳C, we first use the Ideal Gas Law to find the number of moles, and then we divide the mass of the gas by this number to find its molar mass. In our example, the molar mass of the gas is calculated to be 72.05 g/mol.

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Most popular questions from this chapter

Consider the apparatus shown in the following drawing. (a) When the valve between the two containers is opened and the gases allowed to mix, how does the volume occupied by the \(\mathrm{N}_{2}\) gas change? What is the partial pressure of \(\mathrm{N}_{2}\) after mixing? (b) How does the volume of the \(\mathrm{O}_{2}\) gas change when the gases mix? What is the partial pressure of \(\mathrm{O}_{2}\) in the mixture? (c) What is the total pressure in the container after the gases mix?

Which assumptions are common to both kinetic-molecular theory and the ideal- gas equation?

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