What is the freezing point of an aqueous 1.00 m NaCl solution? (Kf = 1.86 *C/m) (Assume complete dissociation of the salt.)
I know the answer is -3.78*C but I cannot figure out how to get it. I keep getting -1.86*C. Please show step by step so I can understand. Thank you.
Solution:
Given molality of NacL = 1.00 Molal
I= Vant hoff Factor(1=2)
number of Ions present
NaCL= 1Na+ + 1Cl- therefore i=2
ΔTf = Kf*m*i = 1.86 * 1 * 2 = 3.72 degree C
therefor the freezing point of the solution will be (0 degree C - 3.72 degree C ) = -3.72 degree C
0 degree C is the freezing point of water.
therefore -3.72 degree C is the freezing point of the solution.
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