Question

calculate to three significant digits the density of chlorine pentafluoride gas at exactly 5°c and exactly 1 atm. you can assume chlorine pentafluoride gas behaves as an ideal gas under these conditions.

Explanation:

Step1: Convert temperature to Kelvin

The formula for converting Celsius to Kelvin is \(T = t+273.15\). Given \(t = 5^{\circ}C\), then \(T=(5 + 273.15)\text{K}=278.15\text{K}\).

Step2: Find the molar mass of \(ClF_5\)

The molar mass of \(Cl\) is \(M_{Cl}=35.45\text{g/mol}\), and the molar mass of \(F\) is \(M_{F}=19.00\text{g/mol}\). For \(ClF_5\), \(M=(35.45 + 5\times19.00)\text{g/mol}=(35.45+95.00)\text{g/mol}=130.45\text{g/mol}\).

Step3: Use the ideal - gas - based density formula

The ideal gas law is \(PV = nRT\), where \(n=\frac{m}{M}\) (mass \(m\) and molar mass \(M\)). Substituting \(n\) into the ideal gas law gives \(PV=\frac{m}{M}RT\). Rearranging for density \(
ho=\frac{m}{V}\), we get \(
ho=\frac{PM}{RT}\).
Given \(P = 1\text{atm}\), \(R=0.0821\frac{\text{L}\cdot\text{atm}}{\text{mol}\cdot\text{K}}\), \(M = 130.45\text{g/mol}\), and \(T = 278.15\text{K}\).
Substitute the values into the formula: \(
ho=\frac{1\text{atm}\times130.45\text{g/mol}}{0.0821\frac{\text{L}\cdot\text{atm}}{\text{mol}\cdot\text{K}}\times278.15\text{K}}\).
First, calculate the denominator: \(0.0821\times278.15 = 22.836115\).
Then, \(
ho=\frac{130.45}{22.836115}\text{g/L}\approx5.71\text{g/L}\).

Answer:

\(5.71\frac{\text{g}}{\text{L}}\)