chem hw help plz 1.Calculate the freezing point for a 1.71 m solution of CCl 4 i

Question

chem hw help plz

1.Calculate the freezing point for a 1.71 m solution of CCl4 in benzene.

2.What is the osmotic pressure in atm of a 0.151 M solution of NaCl at 0oC?

the answer to question 2 is not 3.38. this is what they say in the feedback...You must use the molarity of particles. Remember NaCl is a strong electrolyte. This is equivalent to ? = (Mi)RT where i is the van't Hoft factor.

Because NaCl is a strong electrolyte it breaks apart into Na+ ions and Cl-when it dissolves in water. Thus each NaCl produces two particles when dissolved in water and the van't Hoff factor is 2.

Explanation / Answer

1)kf for benzene =5.12

so freezing point depression = 1.71*5.12
=8.75

freezing point of benzene=5.5 C

so f=5.5-8.75
=-3.25 C

2) osmotic pressure = CRT*i

=0.151*(1/12)*273*2

=6.87 atm

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