A solution of an organic compund is prepared by dissolving 34.2 g in 500 g of water.Calculate the molecular mass of the compound and freezing point of the solution . Given that `k_(b)` for water = 0.52 `K m^(-1)` , b.pt of solution ` = 100.14^(@)C` , `K_(f)` for water = `1.87 Km^(-1)` .

Answer

`342, 0 .374^(@)C`

Explanation

`Delta T_(B) = 100.104 - 100 = 0.104^(@)`
`M_(B) = (0.52 xx 34.2 xx 1000)/(0.14 xx 500) = 342`
`Delta T_(f) = (1.87 xx 34.2 xx 293)/(342 xx 500) = 0.374`
`therefore ` Freezing point `= - 0.374^(@)C`