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A solution is made by dissolving 0.749 mol of nonelectrolyte solute in 861 g of benzene....

A solution is made by dissolving 0.749 mol of nonelectrolyte solute in 861 g of benzene. Calculate the freezing point, Tf, and boiling point, Tb, of the solution. Constants may be found here.

Solvent Formula Kf value*

(°C/m)

Normal freezing

point (°C)

 Kb value

(°C/m)

Normal boiling

point (°C)
water H2O 1.86 0.00 0.512 100.00
benzene C6H6 5.12 5.49 2.53 80.1
cyclohexane C6H12 20.8 6.59 2.92 80.7
ethanol C2H6O 1.99 –117.3 1.22 78.4
carbon
tetrachloride  		 CCl4 29.8 –22.9 5.03 76.8
camphor   C10H16O 37.8 176
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Answer #1

0.749 mol - 0.870 mol/kg Molality of solution - 0.870 m 861g1000 g Molal depression constant of benzene,k-5.12 °C/m Depreassion in freezing point AT - k, x m 5.12x[0.870 = 4.45 °C Freezing point of solution, Tsoh·berane sort-AZ 5.49°C-4.45°C 1.04°C Molality of solution 0.870 m Molal elevation constant of benzene, t 2.53°Clm According to the elevation in boiling point, A^- k, x m . (2.530cm)(0.870 m)-220°C 00° C Boiling point of water solvent, Tetrsort-1 Elevation in boilingpoint, A7 hter sodment Boiling point of soluti on, ToAIwater solrent = 2.200 C +80.1 0C -82.3°C

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