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The amount (in grams) of sucrose (mol.wt...

The amount (in grams) of sucrose (mol.wt. = 342g) that should be dissolved in 100 g water in order to produce a solution with a `105.0^@C` difference between the boiling point and freezing point is (Given that `k_f=1.86Kkgmol^(-1) and k_b=0.52Kkgmol^(-1)" for water")` Report your answer by rounding it up to to the nearest whole number.

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How many grams of sucrose (molecular weight 342) should be dissolved in 100 g water in order to produce a solution with 105^(@)C difference between the freezing point and the boiling point ? (K_(b) =0.51^(@)C m^(-1) , (K_(f) =1.86^(@)C m^(-1))

How many moles of sucrose should be dissolved in 500 g of water so as to get a solution which has a difference of 104^(@)C between boiling point and freezing point ? ( K_(f) = 1.86 K kg mol^(-1) , K_(b) = 0.52 K kg mol^(-1) )

How many moles of sucrose should be dissolved in 500 g of water so as to get a solution which has a difference of 103.57^(@)C between boiling point and freezing point :- (K_(f)=1.86" K kg mol"^(-1), K_(b)="0.52 K kg mol"^(-1))

The boiling point of an aqueous solution of a non - electrolyte is 100.52^@C . Then freezing point of this solution will be [ Given : k_f=1.86 " K kg mol"^(-1),k_b=0.52 "kg mol"^(-1) for water]

For an aqueous solution freezing point is -0.186^(@)C . The boiling point of the same solution is (K_(f) = 1.86^(@)mol^(-1)kg) and (K_(b) = 0.512 mol^(-1) kg)

The boiling point of an aqueous solution is 100.18 .^(@)C . Find the freezing point of the solution . (Given : K_(b) = 0.52 K kg mol^(-1) ,K_(f) = 1.86 K kg mol^(-1))

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