Answer: The freezing point depression is [tex]1.86^0C[/tex]
Explanation:
Depression in freezing point is given by:
[tex]\Delta T_f=K_f\times m[/tex]
[tex]\Delta T_f[/tex] = Depression in freezing point
[tex]K_f[/tex] = freezing point constant = [tex]1.86^0C/m[/tex]
m= molality = [tex]\frac{\text {moles of solute}}{\text {mass of solvent in kg}}=\frac{2mol}{2kg}=1m[/tex]
[tex]\Delta T_f=1.86mol/kg^0C\times 1m[/tex]
[tex]\Delta T_f=1.86^0C[/tex]
Thus freezing point depression is [tex]1.86^0C[/tex]
