calculate the amount of KCL which must be added to 1Kg of water so that the freezing point is depressed by 2K.(Kf for water =1.86 K Kg/mol)?
Dear student!
We know that depression in freezing point,
ΔTf = Kf x m
Where m = molality i.e no. of moles of solute dissolved in per Kg solvent.
Here ,m = ΔTf /Kf = 2/1.86 = 1.07
We know that molality of KCl (m) = no. of moles of KCl /solvent in Kg.
So, for 1 Kg of water, no . of moles of KCl = 1.07 x 1= 1.07 moles.
Hence, 1.07 moles of KCl is dissolved in the solution.