If 3.2 g of sulphur is heated to a temperature, the sulphur vapour produced occupies a volume of 780 mL at 723 mm pressure and `450^(@)C`. What is the molecular formula of sulphur at this state?

Explanation

Let the molecular formula of sulphur is `S_(x)`.
`therefore 3.2` g of sulphur`=(3.2)/(32xx x)=(1)/(10x)` mol of sulphur
`PV=nRT`
or, `(723)/(760)xx780xx10^(-3)=(1)/(10xx x)xx0.0821xx(450+273)`
`therefore`The molecular formula of sulphur=`S_(8)`.