A solution is prepared by dissolving `1.5g` of a monoacidic base into `1.5 kg` of water at `300K`, which showed a depression in freezing point by `0.165^(@)C`. When `0.496g` of the same base titrated, after dissolution, required `40 mL` of semimolar `H_(2)SO_(4)` solution. If `K_(f)`of water is `1.86 K kg mol^(-1)`, then select the correct statements (s) out of the following( assuming molarity `=` molarity):

Options

A. The `pH` of the solution of weak base is `12.9`.
B. The ionisation constant of the base is `8 xx 10^(-3)`.
C. The osmotic pressure of the aqueous solution of base is `21.67atm`
D. The base is `10%` ionized in aqueous solution.

Answer

A, B, C, D

Explanation

Calculation of molecular weight of `BOH`:
`Eq of base = Eq of H_(2)SO_(4)` used
`(0.469g)/("Mw of base") xx1 = 40 xx 10^(-3) xx (1)/(2) xx2 ("n factor")`
`:. Mw fo BOH = 12.4 g mol^(-1)`
Now: `Delta_(f)T = iK_(f)m`
`0.165 = i xx 1.86 xx (1.5)/(1.24) xx(1000)/(150) = i xx 1.86 xx 0.8` (So `m = 0.8`)
`:. i = (0.165 xx 12.4)/(1.86) = 1.1`
For weak base: `1+ alpha = 1.1, rArr alpha = 0.1%` (Answer `d`)
Thus, `[overset(Theta)OH] = Calpha = 0.8 xx 0.1 = 8 xx 10^(-2)`,
`:. pOH = 1.1, pH = 12.9` (Answer `a`)
So, `K_(b) =C alpha^(2) = 0.8 xx (0.1)^(2) = 8 xx 10^(-3)`, (Answer b)
Also, `pi = iMRT`
`= 1.1 xx 0.8 xx 0.0821 xx 300`
=21.67 atm (Answer c)