Freezing point of a biological fluid is `-0.62^(@)C` in aqueous solution `K_(f)(H_(2)O)=1.86^(@)C.mol^(-1).kg`. The osmotic pressure of the solution at `300K` is (assuming molarity `=` molality)

Calculate osmotic pressure of 0.1M urea aqueous solution at 300K .

If the freezing point of 0.1 M HA(aq) solution is -0.2046^(@)C then pH of solution is ( If K_(f) water =1.86mol^(-1)kg^(-1))

0.2 molal acid HX is 20% ionised in solution, K_(f)=1.86K "molality"^(-1) . The freezig point of the solution is

The freezing poing of an aqueous solution of a non-electrolyte is -0.14^(@)C . The molality of this solution is [K_(f) (H_(2)O) = 1.86 kg mol^(-1)] :

The freezing point of a 0.05 molal solution of a non-electrolyte in water is: ( K_(f) = 1.86 "molality"^(-1) )

An aqueous solution freezes at -0.186^(@)C (K_(f)=1.86^(@) , K_(b)=0.512^(@) . What is the elevation in boiling point?

Mole fraction of a non-electrolyte in aqueous solution is 0.07 . If K_(f) is 1.86^(@) "mol"^(-1)kg , depression if f.p.,DeltaT_(f), is:

The depression in the freezing point of a sugar solution was found to be 0.402^(@)C . Calculate the osmotic pressure of the sugar solution at 27^(@)C. (K_(f) = 1.86" K kg mol"^(-1)) .

Density of 1M solution of a non-electrolyte C_(6)H_(12)O_(6) is 1.18 g/mL . If K_(f) (H_(2)O) is 1.86^(@) mol^(-1) kg , solution freezes at :

A solution is prepared by dissolving 1.5g of a monoacidic base into 1.5 kg of water at 300K , which showed a depression in freezing point by 0.165^(@)C . When 0.496g of the same base titrated, after dissolution, required 40 mL of semimolar H_(2)SO_(4) solution. If K_(f) of water is 1.86 K kg mol^(-1) , then select the correct statements (s) out of the following( assuming molarity = molarity):