What molality of NaCl is required to lower the freezing point of water to -11.4 ⁰C? Use 1.853 ⁰C/m for Kf of water?
Post-lab for Gen Chem lab and can t figure where my prof answer came from. It should be 138.70g but I am completely lost.
Update:Sorry the answer was 3.1m not 138g. Was looking at wrong question.
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Freezing point depression....
ΔT = (i)(Kf)(m)
m = ΔT / (i)(Kf)
m = 11.4C / (2) / 1.853 C/m ............ i = 2 since NaCl --> Na+ + Cl- .... two particles
m = 3.08 m
Based on the number of significant digits in the temperature, the molality can only be expressed to three significant digits.
ΔT = i kf m,
where ΔT is the drop in freezing point,
i is the van’t Hoff factor (moles of dissolved particles (or ions) per mole of solute),
kf is the molal freezing point depression constant,
and m is the molality of the solution.
0 − -11.4 = (2) (1.853) m
m = 3.076 mol/kg