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The addition of 3g of a substance to 100...

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)`

Text Solution

Verified by Experts

The correct Answer is:
(a) `3.816k`, (b) `01814`, (c) `4.669 atm` (d) `250 g mol^(1-)`
(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`
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