The complete question is as follows: If 1.04 g of chlorine gas occupies a volume of 872 mL at a particular temperature and pressure, what volume will 2.08 g of chlorine gas occupy under the same conditions ?
Answer: A volume of 1744 L will 2.08 g of chlorine gas occupy under the same conditions.
Explanation:
Given: [tex]Mass_1[/tex] = 1.04 g, [tex]V_{1}[/tex] = 872 mL
[tex]Mass_2[/tex] = 2.08 g, [tex]V_{2}[/tex] = ?
As molar mass of chlorine is 35.5 g/mol.
Number of moles is the mass of a substance divided by its molar mass.
Hence, moles of chlorine present in 1.04 g chlorine gas is calculated as follows.
[tex]Moles = \frac{mass}{molar mass}\\= \frac{1.04 g}{35.5 g/mol}\\= 0.029 mol[/tex]
Also, moles of chlorine present in 2.08 g chlorine gas is calculated as follows.
[tex]Moles = \frac{2.08 g}{35.5 g/mol}\\= 0.058 mol[/tex]
Formula used to calculate the volume occupied by 2.08 g of chlorine gas is as follows.
[tex]\frac{V_{1}}{n_{1}} = \frac{V_{2}}{n_{2}}[/tex]
Substitute the values into above formula as follows.
[tex]\frac{V_{1}}{n_{1}} = \frac{V_{2}}{n_{2}}\\\frac{872 L}{0.029 mol} = \frac{V_{2}}{0.058 mol}\\V_{2} = 1744 L[/tex]
Thus, we can conclude that a volume of 1744 L will 2.08 g of chlorine gas occupy under the same conditions.