A certain aqueous solution boils at `100.303^@C`. What is its freezing point ?
`K_b` for water = 0.5 `"mol"^(-1)` and `K_f = 1.87` K `"mol"^(-1)`

An aqueous solution boils at 101^(@)C . What is the freezing point of the same solution? (Gives : K_(f) = 1.86^(@)C// m "and" K_(b) = 0.51^(@)C//m )

Some ethylene glycol HOCH_(2)CH_(2)OH is added to your car cooling system along with 5 kg of water .If the freezing point of water glycol solution is -15.0^(@)C what is the boiling point of the solution ? (k_(b)=0.52 kg mol^(-1) and k_(f)=1.86 kg mol^(-1) for water)

A solution of urea in water has a boiling point of 100*18^@C . Calculate the freezing point of the solution.(K for water is 1*86 " K kg " mol^(-1) and K_b " for water is " 0*512 " K kg " mol^(-l) ).

An aqueous solution of NaCI freezes at -0.186^(@)C . Given that K_(b(H_(2)O)) = 0.512K kg mol^(-1) and K_(f(H_(2)O)) = 1.86K kg mol^(-1) , the elevation in boiling point of this solution is:

An aqueous solution contains 5% by weight of urea and 10% by weight of glucose. What will be its freezing point ? ( K_(f) for water =1.86^(@)" mol"^(-) kg)

A solution of urea in water has boiling point of 100.15^(@)C . Calculate the freezing point of the same solution if K_(f) and K_(b) for water are 1.87 K kg mol^(-1) and 0.52 K kg mol^(-1) , respectively.

The freezing point of a 1.00 m aqueous solution of HF is found to be -1.91^(@)C . The freezing point constant of water, K_(f) , is 1.86 K kg mol^(-1) . The percentage dissociation of HF at this concentration is:-

An aqueous solution of 0.1 molal concentration of sucrose should have freezing point (K_(f)=1.86K mol^(-1)kg)

How many moles of sucrose should be dissolved in 500gms of water so as to get a solution which has a difference of 104^(@)C between boiling point and freezing point. (K_(f)=1.86 K Kg "mol"^(-1).K_(b)=0.52K 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)