1.

Calculate the freezing point of an aqueous solution containing 10.50 g ofMgBr_(2) in 200 g of water . (Molar mass of MgBr_(2) = 184 , K_(f) = 1.86 Kkg mol^(-1))

Answer»

Solution :`m = (n_(g) xx 1000)/(W_(A) (g))` <BR> `= (W_(B) xx 1000)/(M_(B) xx W_(A)) = (10.50 xx 1000)/(184 xx 200) = 0.2853` M
`MgBr_(2)` IONIZES as `MgBr_(2) to Mg^(2+) + 2 Br^(-)`
`i= 3`
`DeltaT_(F) = ixx K_(f)xx M`
`= 3 xx 1.86 xx 0.2855`
`=1.59`
Freezing point = `0 - 1.59^(@) C = -1.59^(@) C`


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