1. Solution made by dissolving 6.73 grams of toluic acid in 119 grams of henzene

Question

1. Solution made by dissolving 6.73 grams of toluic acid in 119 grams of henzene (f.p = 5.45°C, Kf =5.07°C/m) gave a freezing point of 3.00°C. What is the calculated molecular weight of toulic acid from this data? (I tried doing it but according to the answers it's wrong and I guess w the -iKfm the negative aspect confused me, can I get a clarification?)

Show work- Please write your complete work in space provided. Partial credit will be given. 1. Solution made by dissolving 6.73 grams of toluic acid in 1 10 grams of benzene ( fp.-5.45 C, Kf-507 p. 557 "C/m) gave a freezing point of 3.00 °C. What is the calculated molecular weight of toluic acid form this data? F5.07°c/m 2.45C 0.41 m bennoe kg benzern 6, 73 0.0528 mol

Explanation / Answer

dTf = i*Kf*m

DTf = 5.45-3

i = vanthoff factor of toluic acid = 1

Kf of benzene = 5.07 c/m

   m = molality = ?

   5.45-3= 1*5.07*m

m = 0.483

molality(m) = (w/Mwt)*(1000/Wt of benzene)

         0.483 = (6.73/x)*(1000/110)

x = Molarmass of toluicacid = 126.67 g/mol

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