Find boiling point (in k) of an 1m aqueous solution of an electrolyte `A_2B_3` WHICH IS 60% ionised.
[Given K_b(H_2O) = 0.52 (K Kg)/mole]

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) )

3.0 molal aqueous solution of an electrolyte A_(2)B_(3) is 50% ionised. The boilng point of the solution at 1 atm is: [k_(b) (H_2O) = 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)

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

What is the boiling point of 1 molal aqueous solution of NaCl (K_b)=0.52 K molal^-1]

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

2.0 molal aqueous solution of an electrolyte X_(2) Y_(3) is 75% ionised. The boiling point of the solution a 1 atm is ( K_(b(H_(2)O)) = 0.52K kg mol^(-1) )

The boiling point of an aqueous solution is 100.18 .^(@)C . Find the freezing point of the solution . (Given : K_(b) = 0.52 K kg mol^(-1) ,K_(f) = 1.86 K kg mol^(-1))

The boiling point of an aqueous solution of a non - electrolyte is 100.52^@C . Then freezing point of this solution will be [ Given : k_f=1.86 " K kg mol"^(-1),k_b=0.52 "kg mol"^(-1) for water]