# chemistry

i need to know if I am in the right tract please let me know...I am calulating the molality of the following exercise so I could be able to apply the formula T=Kf * molality

calculate the freezing point of a solution containin 12.2 grams of benzoic acid, dissolved in 250 grams of nitrobenzene. The freezing point of nitrobenzene is 5.7 C, and its freezing point depression constant is 7.O C/m

M= # mols/#kg solvent

so,

C6H5CO2H

C= 12.011 * 6 = 72.066
H= 1.00794
c= 12.011
O= 15.9994
H= 1.000794
amu= 102.09

12.2 g * 1mol/102.09g =0.120 mol

M= 0.120 mol/0.250Kg= 0.48

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3. 👁 99
1. I didnt' check the math on the molmass, so double check that , but you have it down right procedure.

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2. thansk

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posted by Joshua
3. from the texbook K(f)= 5.12 C/m but from the exercise is 7.0 C/m which one should i use to solve fro the freezing point? please help

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posted by Joshua
4. if so then the solution will be

Delta T= Kf * M

T= 7.0 C/m * O.474mol/g
T=3.32 C

The freezing point of the solution will be

5.7 C - 3.32 C = 2.38 C please let me know if this is correct? please

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posted by Joshua
5. I didn't check math, but it exactly the right way to work it. The symbol for molality is m, and for molarity M. The units for molality are moles/kg

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6. oh i see m ok thanks

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posted by Joshua

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