The boiling point of benzene is 353.2 K. When 1.8 g of a non-volatile solute was dissolved in 90g benzene the boiling point was raised to 354.1 K. Calculate the molecular mass of the solute.
(`K_(b)` of benzene = `2.53 K kg mol^(-1)`)

Explanation

Given :
`T_("benzene") = 353.2 K`
`W_(A) = 1.8 g`
`W_(A) = 90 g`
`T_("solution") = 354.1K`
`K_(b) =2.53 K kg mol^(-1)`
`M_(B) = ?`
We know that, `Delta T_(b) = K_(b)m`
`354.1 - 352.2 = (2.53 xx 1.8 xx 1000)/(M_(B) xx 90)`
`0.9 xx M_(B) xx 90 = 2.53 xx 1.8 xx 1000`
`M_(B) = (2.53 xx 1.8 xx 1000)/(0.9 xx 90) = 56g`