Calculate the freezing point of an aquteous containing 10.50 g of `MgBr_(2)` in 200 g of water (molar mass of `MgBr_(2)=184 mol^(-1),K_(f) for water=1.86 K kg mol^(-1)`)

Answer

`-1.59^(@)C`

Explanation

For `MgBr_(2)`
`MgBr_(2) to Mg^(2+) + 2Br^(-)`
`I = 3`
`Delta T_(f) = (ixxk_(f) xx w_(B) xx 1000)/(w_(A) xx M_(B))`
` =(3 xx 1.86 xx 10.50 xx 1000)/(200 xx 184)`
` = 1.59`
Freezing point ` = 0- 1.59^(@)C`
` = - 1.59^(@)C`