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 molar mass ` M_(B)` is calculated as :
`M_(B) = (k_(f) xx w_(B) xx 1000)/(w_(A) xx Delta T_(f))`
`Delta T_(f) = 0.112^(@)`
`w_(A) =226 g`
`w_(B) = 3.26 g`
`k_(f) = 4.90 km^(-1)`
`M_(B) = (4.90 xx 3.26 xx 1000)/(226 x 0.112)`
` = 632 = x 78.8`
`632 = xx 78.8`
` x = (632)/(78.8) = 8`
`therefore ` Molecular formula of selenium = `Se_(g)`.