1Kg of an aqueous solution of Sucrose is...

1Kg of an aqueous solution of Sucrose is cooled and maintained at `-4^(@)"C".` How much ice will be seperated out if the molality of the solution is 0.75? `"K"_("F")("H"_(2)"O")=1.86"Kg mol"^(-1)"K"`

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1Kg of an aqueous solution of sucrose is cooled and maintained at -4^@C . How much ice will be separated out of the solution if molality of the solution is 0.75. K_f = 1.86

1000 g of 1 molal aqueous solution of sucrose is cooled and maintained at -3.534^(@)C . Find out how much ice will separate out at this temperature. (K_(f) for water =1.86 k m^(-1) )

Elevation in boiling point of an aqueous solution of a non-electroluyte solute is 2.01^(@) . What is the depression in freezing point of the solution ? K_(b)(H_(2)O)=0.52^(@)mol^(-1)kg K_(f)(H_(2)O)=1.86^(@)mol^(-1)kg

1000gm of sucrose solution in water is cooled to -0.5^(@)C . How much of ice would be separated out at this temperature, if the solution started to freeze at -0.38^(@)C . Express your answer in gram. (K_(f)H_(2)O=1.86K "mol"^(-1)kg)

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

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