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))`

Explanation

`w_("solvent") = 100 g , w_("sucrose") = ? `
`T_(b) - T_(f) = 105.0^(@)`
Now , `Delta T_(b) = k_(b) m`
Boiling point of solution , `T_(b) = 100 + Delta T_(b)`
or `" " T_(b) = 100+ K_(b) m " "….(i)`
`Delta T_(f) = k_(f) m`
Freezing point solution , `T_(f) = 0 - Delta T_(f)`
`or" " T_(f) = 0 - K_(f) xx m " "......(ii)`
Subtracting eq.(ii) from eq.(i) .
`T_(b) - T_(f) = 100 + k_(b) xx m - (0-k_(f) xx m)`
`105 = 100 + k_(b) xx m xx k_(f) xx m`
` 105 = 100 + 0.512 xx m + 1.86 xx m`
`2.375 m = 5`
` m = (5)/(2.372) = 2.108`
Now, `" " m = (W_("sucrose") xx 1000)/(M_("sucrose") xx W_("solvent")) `
`therefore " "2.108 = (W_("sucrose") xx 1000)/(342 xx 100)`
`therefore" "w_("sucrose") = (2.108 xx 342 xx 100)/(1000) = 72.09 g`