A solution of a non-volatile solute in water freezes at -0.30^(@)C . The vapour pressure of pure water at 298 K is 23.51 mm Hg and K_(f) for water is 1.86 degree//molal . Calculate the vapour pressure of this solution at 298 K .

A sample of water under supply was found to boil at 102^(@)C at normal temperature and pressure. It the water pure ? Will this water freeze at 0^(@)C ? Comment.

Pure water freezes at 273 K and 1 bar. The addition of 34.5 g of ethanol to 500 g of water changes the freezing point of the solution. Use the freezing point depression constant of water as 2 K kg mol^(-1). The figures shown below represent plots of vapour pressure (V.P.) versus temperature (T). [molecular weight of ethanol is 46 g mol^(-1) Among the following, the option representing change in the freezing point 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

A solution of non-volite solute in water freezes at -0.80^(@)C .The vapour pressure of pure water at 298 K is 23.51 mm Hg .and K_(f) for water is 1.86 degree per molal. Calculate the vapour pressure of this solution at 298 K .

Heavy water freezes at

When water freezes the distance between its molecules :

Calculate the freezing point depression expected for 0.0711 m aqueous solution of Na_(2)SO_(4) . If this solution actually freezes at -0.320 ^(@)C , what would be the value of Van't Hoff factor ? ( K_(f) for water is 1.86 ^(@)C mol^(-1) ) .

0.1mole aqueous solution of NaBr freezes at - 0.335^@C at atmospheric pressure , k_f for water is 1.86^@C . The percentage of dissociation of the salt in solution is

How much ethyl alcohol must be added to 1.00 L of water so that the solution will not freeze at -4^(@)F ? K_f of water = 1.86 ^(@)C/m