2 molal solution of a weak acid HA has a freezing point of `3.885^@C`. The degree of dissociation of this acid is _______ `xx10^(-3)` .(Round off to the Nearest Integer).
[Given : Molal depressing constant of water =1.85 K kg `mol^(-1)` Freezing point of pure water =`0^@C` ]

HA (2 molal) has freezing point of -3.885^@C . Find the dgree of dissociation if K_f = 1.85

AB_(2) is 10% dissociated in water to A^(2+) and B^(-) The boiling point of a 10.0 molal aqueous solution of AB_(2) is "________" ""^(@)C. (Round off to the Nearest Integer). [Given : Molal elevation constant of water k _(b) = 0.5 K kg mol ^(-1) boiling point of pure water = 100^(@)C ]

A solute A dimerizes in water. The boiling point of a 2 molal solution of A is 100. 52^@C . The percentage association of A is ............ (Round off to the Nearest Integer). ["Use "K_b " for water "=0.52 "K kg mol"^(-1) " Boiling point of water "=100^@C]

The freezing point of an aqueous solutions is 272 .93 K . Calculate the molality of the solution if molal depression constant for water is 1.86 Kg mol^(-1)

A 0.2 molal aqueous solution of a weak acid HX is 20% ionized. The freezing point of the solution is (k_(f) = 1.86 K kg "mole"^(-1) for water):

In a 0.2 molal aqueous solution of weak acid HX (the degree of dissociation 0.3 ) the freezing point is (given K_(f) = 1.85 K molality^(-1) ):

If 0.2 molal aqueous solution of a weak acid (HA) is 40% ionised then the freezing point of the solution will be ( K_f for water = 1.86^@ C