If Kf = 1.86C/m, how many grams of calcium chloride must be added to 1.0L of water (assume a density of 1.0 g/mL) to lower the freezing point 4.0 C?
delta T = i*Kf*m
You know delta T, i = 3, Kf is known, solve for m.
m = mols/Kg solvent.
Solve for mols
mols = grams/molar mass.
Solve for grams.
is the answer 0.079g or 0.00000646g ? I'm not sure if I've done it correctly.
I don't buy either answer. If you will show your work I will find the error. Note: I suspect the 0.079 answer is the one you want EXCEPT you didn't convert 1000 g solvent to kg.
delta t = i*Kf*m m = mols/kg solvent
4= (3)(1.86)
m= 0.717 (0.717)(1000) = 0.000717mol
0.000717mol/110.98g/mol = 6.46 x 10^-6g
wait is it 79.57g?
Your algebra, for want of a better word, is a little sloppy. I would not cut so many corners; take that extra time to write out the full equation.
dT = i*K*m
4 = 3*1.86*m
m 4/(3*1.86) = 0.717
m = mols/kg solvent
0.717 = mols/1 kg
mol = 0.717
mols = grams/molar mass
0.717 x 110.98 = 79.55g CaCl2 which I would round to 79.6 g to 3 significant figures..