The addition of 3g of a substance to 100g C CI_(4)(M = 154 g mol^(-1)) raises the boiling point of C CI_(4) by 0.60^(@)C if K_(b)(C CI_(4)) is 5K mol^(-1) kg, calculate (a) the freezing point depression (b) the relative lowering of vapour pressure (c) the osmotic pressure at 298K and (d) the molar mass of the substance. Given: K_(f) (C CI_(4)) = 31.8 K kg mol^(-1) and rho (solution) = 1.64 g cm^(-3)

The addition of 3g of a substance to 100g C CI_(4)(M = 154 g mol^(-1))

1.	The addition of 3g of a substance to 100g C CI_(4)(M = 154 g mol^(-1)) raises the boiling point of C CI_(4) by 0.60^(@)C if K_(b)(C CI_(4)) is 5K mol^(-1) kg, calculate (a) the freezing point depression (b) the relative lowering of vapour pressure (c) the osmotic pressure at 298K and (d) the molar mass of the substance. Given: K_(f) (C CI_(4)) = 31.8 K kg mol^(-1) and rho (solution) = 1.64 g cm^(-3)
Answer» Solution :(a) `(DeltaT_(f))/(DeltaT_(b)) = (K_(f))/(K_(b))` `DeltaT_(f) = (31.8)/(5.03) XX 0.6 = 3.79^(@)C` (b) `m = (DeltaT_(b))/(K_(b)) = (0.6)/(5.03) = (x xx1000)/((1-x)154)` `rArr 1-x = 54.44 x` `rArr 55.44 x = 1` `x = (DeltaP)/(P^(@)) = (1)/(5.44) = 0.018` (c) `(1)/(55.44) = ((3)/(M_(1)))/((3)/(M_(1))+(100)/(154))` `rArr (100)/(154) = (166.32)/(M_(1)) M_(1) = 231.5` `M = (0.6)/(5.03)` `(M xx 1000)/(1000 xx 1.64 -M xx 251.5)` `1640 = 8634.83M` `M = 0.19` `p = n//V RT = 0.19 xx 0.0821 xx 298` `=4.65 atm`