Calculate the freezing point of an aqueous soltuion of non-electrolyte having an osmotic pressure `2.0 atm` at `300 K`. (`K'_(f) = 1.86 K mol^(-1) kg` and `S = 0.0821` litre atm `K^(-1) mol^(-1)`)

The freezing point of an aqueous solution of non-electrolyte having an osmotic pressure of 2.0 atm at 300 K will be, (K_(f) "for " H_(2)O = 1.86K mol^(-1)g) and R = 0.0821 litre atm K^(-1) mol^(-1) . Assume molarity and molarity to be same :

The freezing point of an aqueous solution of non-electrolyte having an osmotic pressure of 2.0 atm at 300 K will be, (K_(f) "for " H_(2)O = 1.86K mol^(-1)g) and R = 0.0821 litre atm K^(-1) mol^(-1) . Assume molarity and molality to be same :

Calculate the freezing point of an aqueous solution of a non-electrolyte having an osmotic pressure 2.0 atm at "300 K. Kf = 1.86 k.kg.mol"^(-1.)" R = 0.0821 lit.atm.k"^(-1)" mol"^(-1)

Calculate the freezing point of an aqueous solution of a non-electrolyte having an osmotic pressure 2.0 atm at "300 K. Kf = 1.86 k.kg.mol"^(-1.)" R = 0.0821 lit.atm.k"^(-1)" mol"^(-1)

Calculate the freezing point of an aqueous solution of a non-electrolyte having an osmotic pressure 2.0 atm at "300 K. Kf = 1.86 k.kg.mol"^(-1.)" R = 0.0821 lit.atm.k"^(-1)" mol"^(-1)

Calculate the freezing point of an aqueous solution of a non-electrolyte having an osmotic pressure of 2.0 atm at 300 K. K_f=1.86 k//m, R=0.0821 L "atm" //k//mol .

Calculate the freezing point of an aqueous solution of a non electrolyte having an osmotic pressure at 2 atm at 300K (K_f=1.86Cm^-1, R=0.082 L atm K^-1 mol^-1)