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

[" An aqueous solution of a solute AB has "],[" of "101.08^(@)C" (AB is "100%" ionized in "f" ) "],[" ing point of the solution and freezes "],[180^(@)C" Hence,"AB(K_(b)/K_(r)=0.3)],[" 1) is "100%" ionized at the f.p.of the solutic "]

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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 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 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

A 0.1 molal aqueous solution of a weak acid is 30% ionized. If K_(f) for water is 1.86^(@)C//m , the freezing point of the solution will be.

A 0.1 molal aqueous solution of a weak acid is 30% ionized. If K_(f) for water is 1.86^(@)C//m , the freezing point of the solution will be.