15 g of an unknown molecular substance was dissolved in 450 g of water. The resulting solution freezes at ` - 0 . 34^(@) C ` . What is them molar ass of the substance ? (`K_(f)` for water = 1. 86 K kg `mol^(-1)`)

Answer

`182 . 35 g mol^(-1)`

Explanation

`W_(B) = 15 g, W_(A) = 450 g = 0.45 kg`
`K_(f) = 1.86 K kg mol^(-1)`
`Delta T_(f) = [0 -(-0.34 )]^(@) C `
`= 0.34^(@) C = 0.34 K`
`M_(B) = ?`
`Delta T_(f) = K_(f) xx (W_(B))/(M_(B)) xx (1)/(W_(A) (kg))`
`M_(B) = (K_(f) xx W_(B))/(Delta T_(f) xx W_(A)(kg))`
`= (1.86 K kg mol^(-1) xx (15 g))/( (0. 34 K) xx (0.45 kg))`
`= (186 xx 1500)/( 34 xx 45) g mol ^(-1)`
`= 182 . 35 g mol^(-1)`