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[" An aqueous solution of a solute AB ha...

[" 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)]

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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)

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) )

1.0 molal aqueous solution of an electrolyte "A"_(2)"B"_(3) is 60% ionised. The boiling point of the solution at 1 atm is ("K"_("b"("H"_(2)"O"))=0.52 "K kg mol"^(-1)")

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)

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

An aqueous solution freezes at 1.186^(@)C ( K_(f)=1.86,K_(b)=0.512 ). What is the elevation in boiling point?