Question

1. The molal freezing pointof water is 1.86°C/m(1.86K/m), which can be expressed as 1.86°C • kg H2O/mol solute) after some algebraic rearrangement. This means that a solution of 1 mol cane sugar (342g, over 3/4 lb) dissolvedin 1kgof water should freeze at -1.86°C.

Describe the freezing point drop suggestin number 1.

Answer 1

$T _f ​ = \frac {1000×K _f ​ ×w} {342×1000} = ​1.86K$

$T_f=0-1.86=-1.86K$