1 molal aqueous solution of an electrolyte A₂B₃ is 60% ionised. The boiling point of the solution at 1 atm is-. (Rounded-off to the nearest integer)
[Given K_b for (H₂O) = 0.52 K kg mol^–1]

When 12.2 g of benzoic acid is dissolved in 100 g of water, the freezing point of the solution was found to be
–0.93 °C. The number (n) of benzoic acid molecules associated (assuming 100% association) is:
(K_f(H₂O) = 1.86K kg mol^–1)

When 9.45 g of ClCH₂COOH is added to 500 mL of water, its freezing point drops by 0.5 ºC.
The dissociation constant of ClCH₂ COOH is x × 10^–3. The value of x is:

(Rounded off to the nearest integer)

$[{\mathrm{K}}_{\mathrm{f}\left({\mathrm{H}}_2\mathrm{O}\right)}=\mathrm{1.86<mspace\; width="1em"></mspace>}{\mathrm{kgmol}}^{-1}]$

1. 36 × 10^-3 
2. 29 × 10^-3 
3. 45 × 10^-3 
4. 30 × 10^-3 

Molal depression constant for a solvent is 4.0 K kg mol^–1 . The depression in the freezing point of the solvent for 0.03 mol kg^–1 solution of K₂SO₄ is : (Assume complete dissociation of the electrolyte)

1. 0.36 K
2. 0.18 K
3. 0.12 K
4. 0.24 K

The degree of dissociation $\left(\alpha \right)$ of a weak electrolyte, A_xB_y is related to van't Hoff factor (i) by the expression :

1. $\begin{array}{l}\alpha =\frac{i-1}{(x+y-1)}\end{array}$

2. $\begin{array}{l}\alpha =\frac{i-1}{x+y+1}\end{array}$

3. $\alpha =\frac{x+y-1}{i-1}$

4. $\alpha =\frac{x+y+1}{i-1}$

A 0.03 M aqueous solution of a weak acid HX has an acid dissociation constant \(K_a = 1.2 \times 10^{-5}\), for the dissociation\(\mathrm{HX}_{(\mathrm{aq})} \rightleftharpoons \mathrm{H}_{(\mathrm{aq})}^{+}+\mathrm{X}_{(\mathrm{aq})}^{-} \). Calculate the osmotic pressure (in atm) of the solution at 300 K.