To determine the mass of the gas sample, follow these steps:
1. Convert the volume from milliliters to liters:
- Given volume: 520.6 milliliters
- Conversion factor: 1 liter = 1000 milliliters
[tex]\[
\text{Volume in liters} = \frac{520.6 \, \text{mL}}{1000} = 0.5206 \, \text{L}
\][/tex]
2. Convert the temperature from Celsius to Kelvin:
- Given temperature: 5.9°C
- Conversion from Celsius to Kelvin: [tex]\( \text{Temperature in Kelvin} = \text{Temperature in Celsius} + 273.15 \)[/tex]
[tex]\[
\text{Temperature in Kelvin} = 5.9 + 273.15 = 279.05 \, \text{K}
\][/tex]
3. Use the Ideal Gas Law to find the number of moles:
The Ideal Gas Law is given by: [tex]\( PV = nRT \)[/tex]
- [tex]\( P \)[/tex] is the pressure in atmospheres (atm)
- [tex]\( V \)[/tex] is the volume in liters (L)
- [tex]\( n \)[/tex] is the number of moles
- [tex]\( R \)[/tex] is the Ideal Gas Constant, which is [tex]\( 0.0821 \, \text{L·atm/(K·mol)} \)[/tex]
- [tex]\( T \)[/tex] is the temperature in Kelvin (K)
Given:
- Pressure ([tex]\(P\)[/tex]) = 33.44 atm
- Volume ([tex]\(V\)[/tex]) = 0.5206 L
- Temperature ([tex]\(T\)[/tex]) = 279.05 K
Rearrange the Ideal Gas Law to solve for [tex]\( n \)[/tex]:
[tex]\[
n = \frac{PV}{RT}
\][/tex]
[tex]\[
n = \frac{33.44 \, \text{atm} \times 0.5206 \, \text{L}}{0.0821 \, \text{L·atm/(K·mol)} \times 279.05 \, \text{K}} = 0.75988 \, \text{moles}
\][/tex]
4. Calculate the mass of the gas:
- Molar mass of the gas ([tex]\(M\)[/tex]) = 93 g/mol
The mass ([tex]\(m\)[/tex]) can be calculated using the formula:
[tex]\[
m = n \times \text{molar mass}
\][/tex]
[tex]\[
m = 0.75988 \, \text{moles} \times 93 \, \text{g/mol} = 70.67 \, \text{grams}
\][/tex]
5. Round the mass to the nearest tenth:
The mass rounded to the nearest tenth is:
[tex]\[
70.7 \, \text{grams}
\][/tex]
Therefore, the mass of the gas sample is [tex]\( 70.7 \)[/tex] grams.
