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 grams of sucrose (molecular mass 342) should be dissolved in 100 g water in order to produce a soluton having difference between the boiling point and freezing point equal to 105^@C ? (K_b=0.512 Km^-1, K_f=1.86Km^-1) .

How many grams of sucrose should be dissolved in 100g water in order to produce a solution having difference between the free zing point and the boiling point temperatures at 105.0^@C

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.

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.

'x' gm of solute (M. mass = 342) should be dissolved in 500gm of water so as to get a solution having difference of 105^@C between freezing point and boiling point, then 'x' is

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) )