Chapter 5, Problem 5.36P

Interpretation Introduction

Interpretation:

Mass of $357\text{ mL}$ sample of chlorine trifluoride gas $\left({\text{CLF}}_{\text{3}}\right)$ at $\text{699}\text{mmHg}$ and $45°\text{C}$ is to be calculated.

Concept introduction:

Ideal gas law equation is as follows:

$PV=nRT$ (1)

Here,

$P$ is the pressure of the gas.

$V$ is the volume occupied by gas.

$n$ is the no. of moles of gas.

$T$ is the temperature in Kelvin.

$R$ is the universal gas constant.

The formula to calculate no. of moles  is as follows:

$n=\frac{\text{weight}\left(m\right)}{\text{molecular}\text{weight}\left(M\right)}$ (2)

Conversion factor to convert $\text{ml}$ to $\text{L}$ is as follows:

$1\text{ml}={10}^{-3}\text{L}$

Conversion factor to convert $\text{mmHg}$ to $\text{atm}$ is as follows:

$\text{760}\text{mmHg}=\text{1}\text{atm}$

The conversion factor to convert $°\text{C}$ to Kelvin is as follows:

$T\left(\text{Kelvin}\right)=T\left(\text{°C}\right)+273.15$ (3)