What mass of `NaCI ("molar mass" =58.5g mol^(-1))` be dissolved in `65g` of water to tower the freezing point by `7.5^(@)C`? The freezing point depression constant, `K_(f)`, for water is `1.86 K kg mol^(-1)`. Assume van't Hoff factor for `NaCI` is `1.87`.

Answer

8.20 g

Explanation

Lowering in freezing point `Delta T_(f) = 7.50^(@)C`
`K_(f) = 186^(@)C//m`
Mass of water , `w_(A) = 65.0g`
Molar mass of `NaCl, M_(B) = 58.5`
Van.t Hoff factor , I = 1.87, Mass of NaCl , `w_(B) = ?`
`DeltaT_(f) = (iK_(f) xx 1000 xx w_(B))/(w_(A) xx M_(B))`
or `w_(B) = (Delta T_(f) xx w_(A) xx M_(B))/( i xx K_(f) xx 1000)`
` = (7.50 xx 65.0 xx 58.5)/(1.87 xx 1.86 xx 1000) = 8.20 g`