The freezing point of a `0.50` molal aq. solution of weak acid (HA) is `-1.302^(@)C`. The dissociation constant of acid HA is [Given: `K_(f)` for water `=1.86kgmol^(-1)K]`

The freezing point of a 0.08 molal solution of NaHSO^(4) is -0.372^(@)C . Calculate the dissociation constant for the reaction. K_(f) for water = 1.86 K m^(-1)

The freezing point of a 0.05 molal solution of a non-electrolyte in water is [K_(f)=1.86K//m]

The freezing point of a 1.00 m aqueous solution of HF is found to be -1.91^(@)C . The freezing point constant of water, K_(f) , is 1.86 K kg mol^(-1) . The percentage dissociation of HF at this concentration is:-

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

The freezing point of 0.1 molal aq. Solution of K_(X)[Fe(CN)_(6)] is -0.744^(@) C. The molal depression constant of water is 1.86 K-kg mol^(-1) . If the salt behaves as strong electrolyte in water, the value of X is :

A 0.2 molal aqueous solution of weak acid (HX) is 20% ionised. The freezing point of this solution is (Given, K_(f) = 1.86^(@) C m^(-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) ):

The freezing point depression of a 0.1 M aq. Solution of weak acid (HX) is -0.20^(@)"C". What is the value of equilibrium constant for the reaction: "HX(aq)"iff"H"^(+)"(aq)"+"X"^(-1)"(aq)" [Given: "K"_("f")" for water=1.8 Kg mol"^(-1)"K. & Molality=Molarity" ]