An aqueous solution of a solute AB has b.p of `101.08^(@)C` (AB is `100%` ionised at boiling point of the solution ) and freezes at `-1.80^(@)C` . Hence , AB (`K_(b) //K_(f) = 0.3)`:

An aqueous solution of an electrolyte AB has b.pt of 101.08^(@)C . The solute is 100% ionised ay b.pt of water. The f.pt of the same solution is -1.80^(@)C . Hence AB(K_(b)//K_(f) = 0.3)

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

0.1 molal aqueous solution of an electrolyte AB_(3) is 90% ionised. The boiling point of the solution at 1 atm is ( K_(b(H_(2)O)) = 0.52 kg " mol"^(-1) )

0.1 molal aqueous solution of an electrolyte AB_(3) is 90% ionised. The boiling point of the solution at 1 atm is ( K_(b(H_(2)O)) = 0.52 kg " mol"^(-1) )

AN aqueous solution of a substacne freezes at -0.186^(@)C. The boiling point of the same solution (k _(f) =1.86 Km^(-1), K_(b)=0.512 Km^(-1)) 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)

1.0 molal aqueous solution of an electrolyte X_(3)Y_(2) is 25% ionised . The boiling point of the solution is : ( K_(b) for H_(2)O = 0.52K kg/mol)

A 0.2 Molal aqueous solution of a weak acid (HX) is 20 % ionised . The freezing point of this solution is ______. ("Given " =K_(f) =1.86^(@)C/m for water )