Chapter 8, Problem 96QRT

Interpretation Introduction

Interpretation:

Density of $C{l}_2$ and $S{O}_2$ has to be calculated at $0.750\text{ atm}$ and $25.0°C.$  Also density of $C{l}_2$ at $0.750\text{ atm}$ and $35.0°C$ and density of $S{O}_2$ at $2.60\text{ atm}$ and $25.0°C$ has to be determined.

Concept Introduction:

Density can be calculated by the equation given below:

Density can be calculated by the equation given below:

  $\text{Density}\text{=}\frac{\text{PM}}{\text{RT}}$

Where,

$P$ is the Pressure

$\text{M}$ is the Molar mass

$\text{R}$ is the Gas constant

$\text{T}$ is the Temperature

Answer to Problem 96QRT

• Density of $C{l}_2$ at $0.750\text{ atm}$ and $25.0°C$ is $2.17g/L.$
• Density of $S{O}_2$ at $0.750\text{ atm}$ and $25.0°C$ is $1.96g/L.$
• Density of $C{l}_2$ at $0.750\text{ atm}$ and $35.0°C$ is $2.10g/L.$
• Density of $S{O}_2$ at $2.60\text{ atm}$ and $25.0°C$ is $6.80g/L.$

Explanation of Solution

Given data is shown below,

  $\begin{array}{l}\text{T = 25}°\text{C = }\left(25+273\right)K=\text{ 298K}\\ \\ \text{T = 35}°\text{C = }\left(35+273\right)K=\text{ 308K}\end{array}$

Molar mass of $C{l}_2$ is determined as given,

  $\begin{array}{c}{M}_{C{l}_2}=\left(2\times \text{ 35}\text{.45 g/mol}\right)\\ \\ =\text{ 70}\text{.90 g/mol}\end{array}$

Molar mass of $C{l}_2$ is $\text{70}\text{.90 g/mol}\text{.}$

Molar mass of $S{O}_2$ is determined as given,

  $\begin{array}{c}{M}_{{\text{SO}}_2}\text{ = Atomic mass of S + }\left(2\times \text{ Atomic mass of O}\right)\\ \\ =\left(32.06\text{ g/mol}\right)\text{ +}\left(2\times \text{ 16}\text{.00 g/mol}\right)\\ \\ =\text{ 64}\text{.06 g/mol}\end{array}$

Molar mass of $S{O}_2$ is $\text{64}\text{.06 g/mol}\text{.}$

• Calculate density of $C{l}_2$ at $0.750\text{ atm}$ and $25.0°C$

Density of $C{l}_2$ can be calculated as follows,

  $\begin{array}{c}\text{Density}\text{=}\frac{\text{PM}}{\text{RT}}\\ \\ =\frac{\left(0.750\text{ atm}\right)\left(\text{70}\text{.9 g/mol}\right)}{\left(0.0821\text{ L}\text{.atm}{\text{.mol}}^{-1}.K^{-1}\right)\left(298\text{ K}\right)}\\ \\ =\text{ 2}\text{.17 g/L}\end{array}$

Density of $C{l}_2$ at $0.750\text{ atm}$ and $25.0°C$ is $2.17g/L.$

• Calculate density of $S{O}_2$ at $0.750\text{ atm}$ and $25.0°C$

Density of $S{O}_2$ can be calculated as follows,

  $\begin{array}{c}\text{Density}\text{=}\frac{\text{PM}}{\text{RT}}\\ \\ =\frac{\left(0.750\text{ atm}\right)\left(\text{64}\text{.06 g/mol}\right)}{\left(0.0821\text{ L}\text{.atm}{\text{.mol}}^{-1}.K^{-1}\right)\left(298\text{ K}\right)}\\ \\ =\text{ 1}\text{.96 g/L}\end{array}$

Density of $S{O}_2$ at $0.750\text{ atm}$ and $25.0°C$ is $1.96g/L.$

• Calculate density of $C{l}_2$ at $0.750\text{ atm}$ and $35.0°C$

Density of $C{l}_2$ can be calculated as follows,

  $\begin{array}{c}\text{Density}\text{=}\frac{\text{PM}}{\text{RT}}\\ \\ =\frac{\left(0.750\text{ atm}\right)\left(\text{70}\text{.9 g/mol}\right)}{\left(0.0821\text{ L}\text{.atm}{\text{.mol}}^{-1}.K^{-1}\right)\left(308\text{ K}\right)}\\ \\ =\text{ 2}\text{.10 g/L}\end{array}$

Density of $C{l}_2$ at $0.750\text{ atm}$ and $35.0°C$ is $2.10g/L.$

• Calculate density of $S{O}_2$ at $2.60\text{ atm}$ and $25.0°C$

Density of $S{O}_2$ can be calculated as follows,

  $\begin{array}{c}\text{Density}\text{=}\frac{\text{PM}}{\text{RT}}\\ \\ =\frac{\left(2.60\text{ atm}\right)\left(\text{64}\text{.06 g/mol}\right)}{\left(0.0821\text{ L}\text{.atm}{\text{.mol}}^{-1}.K^{-1}\right)\left(298\text{ K}\right)}\\ \\ =\text{ 6}\text{.80 g/L}\end{array}$

Density of $S{O}_2$ at $2.60\text{ atm}$ and $25.0°C$ is $6.80g/L.$