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

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 ]

A 0.5% (by weight) solution of A_(2)B in solvent C was found to freeze at -3.25^(@)C . Calculate the degree of dissociation of A_(2)B in solvent C into A_(2)^(-) and B_(2)^(-) . (Given freezing point of pure C is -3^(@)C , molar weight of A_(2)B is 60 and K_(f) of C is 2 K^(-1) "molality"^(-1) ).

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) ):

An aqueous glucose solution has boiling point of 374.19K . Another aqueous glucose solution has a freezing point of 272.22K . Find the molality ratio of two solution. Boiling point and freezing point of H_(2)O373.15K and 273.15K respectively.

Depression in freezing point of 0.1 molal solution of HF is -0.201^(@)C . Calculate percentage degree of dissociation of HF. (K_(f)=1.86 K kg mol^(-1)) .

0.15 molal solution of NaCI has freezing point -0.52 ^(@)C Calculate van't Hoff factor . (K_(f) = 1.86 K kg mol^(-1) )

The cryoscopic contant of water is 1.86 K kg mol^(-1) . A 0.01 molal acetic acid solution produces a depression of 0.0194^(@)C in the freezing point. The degree of dissociation of acetic acid is :

An aqueous solution of urea has freezing point of -0.52^(@)C . If molarity and molality are same and K'_(f) for H_(2)O = 1.86 K "molality"^(-1) the osmotic pressure of solution would be: