Chapter 10, Problem 116RQ

Interpretation Introduction

Interpretation:

The molecular mass of the given unknown gas that decomposes into ${\text{O}}_2$ and $\text{S}$, is to be calculated.

Concept Introduction:

The ideal gas equation, which shows the relationship between the pressure, volume, temperature, and the amount of the gas is as follows:

$\frac{PV}{T}\alpha n$

Here, $P$ is the pressure of the gas, $V$ is the volume of the gas, $T$ is the temperature of the gas, and $n$ is the amount of the gas in terms of moles. A constant is introduced for removing the proportionality sign and for calculating the various terms. This sign is called the universal constant, and is symbolically represented as $\text{R}$, thus, the above expression becomes:

$\frac{PV}{T}=\text{R}n$

The percentage of the mass of the given substance is calculated using the formula:

$m\%=\frac{m}{m_{\text{total}}}\times 100$

Here, $m$ is the mass of the substance of which the percentage needs to be calculated and $m_{\text{total}}$ is the total mass of the mixture.

The relation of the number of moles $\left(n\right)$ of a gas with its mass $\left(m\right)$ and molar mass $\left(M\right)$, is given as:

$n=\frac{m}{M}$.

The conversion of pressure from $\text{mm Hg}$ to $\text{torr}$ can be done using the conversion factor:

$\begin{array}{l}\text{1 mm Hg}=1\text{ torr}\\ \frac{1\text{ torr}}{\text{1 mm Hg}}\end{array}$.

The conversion of the unit of pressure, from $\text{torr}$ to $\text{atm}$, is done using the conversion factor:

$\begin{array}{l}1\text{ atm}=760\text{ torr}\\ \frac{1\text{ atm}}{760\text{ torr}}\end{array}$

The conversion of the unit of volume, from milliliter to liter, is done using the conversion factor:

$\begin{array}{l}1\text{ L = 1000 mL}\\ \frac{1\text{ L}}{1000\text{ mL}}\end{array}$