An aqueous solution of urea has freezing point of `-0.52^(@)C`. If molarity and molality are same and `K'_(f)` for `H_(2)O = 1.86 K "molality"^(-1)` the osmotic pressure of solution would be:

An aqueous solution of urea has a freezing point of -0.52^(@)C . Assuming molarity same for the solution, the osmotic pressure of solution at 37^(@)C would be : (K_(f) of H_(2)O=1.86 K molarity .^(-1) )

A 0.025 m solution of mono basic acid had a freezing point of 0.06^(@)C . Calculate K_(a) for the acid K_(f) for H_(2)O=1.86^(@) "molality"^(-1) .

An aqueous solution of urea has a freezing point of -0.515^@C . Predict the osmotic pressure (in atm) of the same solution at 37^@C . k_f = 1.86 k kg (mol)^(-1)

An aqueous solution of urea has a freezing point of -0.515^(@)C . Predict the osmotic pressure (in atm) of the same solution at 37^(@)C . Given: Kf (Water) = 1.86K kg per mol

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

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

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)] :

0.15 molal solution of NaCI has freezing point -0.52 ^(@)C Calculate van't Hoff factor . (K_(f) = 1.86 K kg mol^(-1) )