The freezing point of a solution containing `50 cm^(3)` of ethylene glycol in `50 g` of water is found to be `-34^(@)C`. Assuming ideal behaviour, Calculate the density of ethylene glycol `(K_(f)` for water = `1.86 K kg mol^(-1)`).

Explanation

Depression in freezing point ,
`Delta T_(f) = (k_(f) xx w_(B) xx 1000)/(M_(B) xx w_(A))`
`K_(f) = 1.86 K m^(-1), Delta T_(f) = 0 - (-34) = 34`
`w_(B) = ? , w_(A) = 50 g, M_(B) = 62`
`therefore" "34= (1.86 xx w_(B) xx 1000)/(62 xx 50)`
` therefore " " w_(B) = (34 xx 62 xx 50)/(1.86 xx 1000) = 56.67`
Now , `" "w_(B) = V xx d`
`" "56.67 = 50 xx d`
` or" "d = (56.67)/(50) = 1.13 g cm^(-3)`