Chapter 10, Problem 84E

Interpretation Introduction

Interpretation:

The molar mass of unknown gas with mass $\text{2}\text{.85 g}$, volume $\text{750 mL}$ at $\text{760 mm Hg}$ and $100°\text{C}$ is to be stated.

Concept introduction:

The combination of all three ideal gas laws: Boyle's Law, Charles' Law, and Gay-Lussac's Law gives rise to a combined gas law. According to this law, pressure, volume, temperature and number of moles of a gas are related to each other.

Answer to Problem 84E

The molar mass of unknown gas with mass $\text{2}\text{.85 g}$, volume $\text{750 mL}$ at $\text{760 mm Hg}$ and $100°\text{C}$ is $\text{116}\text{.4}\text{g}/\text{mol}$.

Explanation of Solution

The equation for the combined gas law is given below.

$PV=nRT$ …(1)

Where,

• $P$ is the pressure of unknown gas.

• $V$ is the volume of unknown gas.

• $n$ is the number of moles of unknown gas.

• $T$ is the temperature of unknown gas.

• $R$ is the gas constant.

• $V$ is the volume of unknown gas.

The given value of pressure is $\text{760 mm Hg}$.

The given value of volume is $\text{750 mL}$.

The given value of temperature is $\text{100}°\text{C}$.

The given value of mass is $\text{2}\text{.85 g}$.

The given value of $R$ is $0.0821\text{L}\cdot \text{atm}/\text{mol}\cdot \text{K}$.

The formula of number of moles is shown below.

$n\left(\text{mol}\right)=\frac{\text{Mass of unknown gas}\left(\text{g}\right)}{\text{Molar mass of unknown gas}}$ …(2)

Substitute the value of (2) in equation (1) as shown below.

$\begin{array}{rll}PV& =& nRT\\ PV& =& \left(\frac{\text{Mass of unknown gas}\left(\text{g}\right)}{\text{Molar mass of unknown gas}}\right)RT\end{array}$ …(3)

Conversion of temperature into Kelvin is shown below.

$\begin{array}{rll}T& =& \left(T(°C)+273.15\right)\text{K}\\ & =& (100+273.15)\text{K}\\ & =& 373.15\text{K}\end{array}$

Conversion of volume into liters is shown below.

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

Value of $\text{750}\text{mL}$ in $\text{L}$ is stated below.

$\begin{array}{rll}\text{750}\text{mL}& =& \frac{\text{1 L}}{\text{1000 mL}}\times \text{750 mL}\\ & =& \text{0}\text{.75 L}\end{array}$

Conversion of $\text{760 mm Hg}$ pressure into $\text{atm}$ is shown below.

$\text{760 mm Hg}=\text{1}\text{atm}$

The value of $P,V,R,T$ and mass of unknown gas is $\text{1}\text{atm},0.75\text{ L},0.0821\text{L}\cdot \text{atm}/\text{mol}\cdot \text{K},373.15\text{K}$ and $\text{2}\text{.85 g}$ respectively.

Substitute the values of $P,V,R,T$ and mass of unknown gas in the equation (3) as shown below.

$\begin{array}{rll}PV& =& \left(\frac{\text{Mass of unknown gas}\left(\text{g}\right)}{\text{Molar mass of unknown gas}}\right)RT\\ \text{1 atm}\times \text{0}\text{.75}\text{L}& =& \left(\frac{\text{2}\text{.85 g}}{\text{Molar mass of unknown gas}}\right)\times 0.0821\text{L}\cdot \text{atm}/\text{mol}\cdot \text{K}\times 373.15\text{K}\\ \text{Molar mass of unknown gas}& =& \left(\frac{\text{2}\text{.85 g}}{\text{1 atm}\times \text{0}\text{.75}\text{L}}\right)\times 0.0821\text{L}\cdot \text{atm}/\text{mol}\cdot \text{K}\times 373.15\text{K}\\ \text{Molar mass of unknown gas}& =& \text{116}\text{.4}\text{g}/\text{mol}\end{array}$

Conclusion

The molar mass of unknown gas with mass $\text{2}\text{.85 g}$, volume $\text{750 mL}$ at $\text{760 mm Hg}$ and $100°\text{C}$ is calculated as $\text{116}\text{.4}\text{g}/\text{mol}$.