Question

O GASES Calculating mole fraction in a ges mbxture Christo A 10.00 L tank at 10.4 °C is filled with 3.99 g of sulfur hexafluoride gas and 14.6 g of chlorine pentafluoride gas. You can assume both gases behave as ideal gases under these conditions Calculate the mole fraction of each gas. Round each of your answers to 3 significant digits. mole fraction gas ? X sulfur hexafluoride chlorine pentafluoride

Answer 1

Answer:-

Given:-

volume of tank (V) = 10.00 L

temperature (T) = 10.4 ⁰C

wt. or mass of sulfur hexafluoride gas (SF₆) = 3.99 g

wt. or mass of chlorine pentafluoride gas (ClF₅) = 14.6 g

mole fraction of sulfur hexafluoride gas ($\chi$_SF6) = ?

mole fraction of chlorine pentafluoride gas ($\chi$_ClF5) = ?

As we know that

molar mass of sulfur hexafluoride gas (SF₆) = molar mass of S + 6 $\times$ molar mass of F

molar mass of sulfur hexafluoride gas (SF₆) = 32 + 6 $\times$ 19

molar mass of sulfur hexafluoride gas (SF₆) = 32 + 114

molar mass of sulfur hexafluoride gas (SF₆) = 146 g / mol

similarly

molar mass of chlorine pentafluoride gas (ClF₅) = molar mass of Cl + 5 $\times$ molar mass of F

molar mass of chlorine pentafluoride gas (ClF₅) = 35.5 + 5 $\times$ 19

molar mass of chlorine pentafluoride gas (ClF₅) = 35.5 + 95

molar mass of chlorine pentafluoride gas (ClF₅) = 130.5 g / mol

Also we know that

No. of moles compound (n) = wt.or mass of compound (g) / molar mass of compound ( g / mol)

therefore

No. of moles of sulfur hexafluoride gas (n_SF6) = wt.or mass of sulfur hexafluoride gas (SF₆) (g) / molar mass of sulfur hexafluoride gas (SF₆) ( g / mol)

No. of moles of sulfur hexafluoride gas (n_SF6) = 3.99 g / 146 g / mol

No. of moles of sulfur hexafluoride gas (n_SF6) = 0.0273 mol

similarly

No. of moles of chlorine pentafluoride gas (n_ClF5) = wt.or mass of chlorine pentafluoride gas (ClF₅) (g) / molar mass of chlorine pentafluoride gas (ClF₅) ( g / mol)

No. of moles of chlorine pentafluoride gas (n_ClF5) = 14.6 g / 130.5 g / mol

No. of moles of chlorine pentafluoride gas (n_ClF5) = 0.1119 mol

So

Total no. of moles present in 10.00 L tank (n_total) = No. of moles of sulfur hexafluoride gas (n_SF6) + No. of moles of chlorine pentafluoride gas (n_ClF5)

Total no. of moles present in 10.00 L tank (n_total) = 0.0273 mol + 0.1119 mol

Total no. of moles present in 10.00 L tank (n_total) = 0.1392‬ mol

therefore

According to the formula

mole fraction of sulfur hexafluoride gas ($\chi$_SF6) = No. of moles of sulfur hexafluoride gas (n_SF6) / Total no. of moles present in 10.00 L tank (n_total)

mole fraction of sulfur hexafluoride gas ($\chi$_SF6) = 0.0273 mol / 0.1392‬ mol

mole fraction of sulfur hexafluoride gas ($\chi$_SF6) = 0.1961

mole fraction of sulfur hexafluoride gas ($\chi$_SF6) = 0.196 ( 3 significant digits) i.e the answer

similarly

mole fraction of chlorine pentafluoride gas ($\chi$_ClF5) = No. of moles of chlorine pentafluoride gas (n_ClF5) / Total no. of moles present in 10.00 L tank (n_total)

mole fraction of chlorine pentafluoride gas ($\chi$_ClF5) = 0.1119 mol / 0.1392‬ mol

mole fraction of chlorine pentafluoride gas ($\chi$_ClF5) = 0.8038

mole fraction of chlorine pentafluoride gas ($\chi$_ClF5) = 0.804 ( 3 significant digits) i.e the answer