An aqueous solution contains `6%` and `4%` by mass of `X` (mol.wt.`=120`) and `Y` (mol.wt.`=180`) respectively. The boiling point of solution `=` `(K_(b)=0.52K.kg.mol^(-1))`

An aqueous solution freezes on 0.36^@C K_f and K_b for water are 1.8 and 0.52 k kg mol^-1 respectively then value of boiling point of solution as 1 atm pressure is

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) )

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) )

In a 0.5 molal solution KCl, KCl is 50% dissociated. The freezing point of solution will be ( K_(f) = 1.86 K kg mol^(-1) ):

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) :

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)]

For an aqueous solution freezing point is -0.186^(@)C . The boiling point of the same solution is (K_(f) = 1.86^(@)mol^(-1)kg) and (K_(b) = 0.512 mol^(-1) kg)

An aqueous solution containing 21.6 mg at a solute in 100 ml of solution has an osmotic pressure of 3.70 mm of Hg at 25^(@)C . The molecular wt of solute in g//mol is

A 5 per cent aqueous solution by mass of a non-volatile solute boils at 100.15^(@)C . Calculate the molar mass of the solute. K_(b)=0.52 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]