The boiling point of an aqueous solution is 374.4 K. What will be its freezing point?(K_(f) = 1.86 K kg mol^(-1) , K_(b) ...

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))

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 500 g 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) = 0.52 K kg mol^(-1) )

The boiling point of an aqueous solution of a non-volatile solute is 100.15^(@)C . What is the freezing point of an aqueous solution obtained by dilute the above solution with an equal volume of water. The values of K_(b) and K_(f) for water are 0.512 and 1.86^(@)C mol^(-1) :

An aqueous solution freezes at 272.4 K while pure water freezes at 273 K. Given K_(f)=1.86 K kg "mol"^(-1) , K_(b)=0.512 K kg "mol"^(-1) and vapour pressure of water at 298 K = 23.756 mm Hg. Determine the following. Boiling point of the solution is

An aqueous solution freezes at 272.4 K while pure water freezes at 273 K. Given K_(f)=1.86 K kg "mol"^(-1) , K_(b)=0.512 K kg "mol"^(-1) and vapour pressure of water at 298 K = 23.756 mm Hg. Determine the following. Depression in freezing point of solution