The molal freezing point depression constant of benzene`(C_(6)H_(6))` is `4.90 K kg mol^(-1)`. Selenium exists as a polymer of the type `Se_(x)`. When `3.26 g` of selenium is dissolved in `226 g` of benzene, the observed freezing point is `0.112^(@)C` lower than pure benzene. Deduce the molecular formula of selenium. (Atomic mass of `Se=78.8 g mol^(-1)`)

Explanation

The given values are:
`K_(f)= 4.90 K kg mol^(-1)`
`W_(solute)= 3.26 g `
`W_(solvent)= 226 g `
`DeltaT_(f)= 0.112^(@)C `
Now, molecular mass can be calculated as
`Mw_(solute)=(K_(f)xxW_(solute)xx1000)/(W_(solvent)xxDeltaT_(t))`
`=(4.901xx3.26xx1000)/(226xx0.112)=632 g mol^(-1)`
Now, molecular mass of `Se_(x) = x xx78.8`
`632=x xx78.8`
`or x=632/78.8=8`
`:.` Molecular formula of selenium =`Se_(8)`.