Question

Please write an explanation if you find the answer, I don't get it.

A flask contains 21.8 g of chlorine gas and 47.8 g of sulfur dioxide gas. What is the mole fraction of the sulfur dioxide?

Round your answer to 3 decimal spaces, even if sig figs are not conserved.

Answer 1

The mole fraction of the sulfur dioxide gas present in a falsk which contains 21.8 g of chlorine gas and 47.8 g of sulfur dioxide gas is 0.708.

How do we calculate mole fraction?

Mole fraction of any substance will be calculated by dividing the moles of desired substance by the total moles of the species present in that sample.

Moles can be calculated as:

n = W/M, where

W = given mass

M = molar mass

Moles of 21.8g of chlorine gas = 21.8g / 71g/mol = 0.307mol

Moles of 47.8g of sulphur dioxide gas = 47.8g / 64g/mol = 0.746mol

Mole fraction of sulphur dioxide gas = 0.746 / 0.746+0.307 = 0.708

Hence required mole fraction of sulphur dioxide gas is 0.708.

To know more about mole fraction, visit the below link:
https://brainly.com/question/1601411

#SPJ1