## Question

###### Part c calculate the freezing/boiling point for 18.0 g of decane, C10H22, in 50.0 g CHCl3...

part c

calculate the freezing/boiling point for 18.0 g of decane, C10H22, in 50.0 g CHCl3

part e

calculate the freezing/boiling point for 0.48 mol ethylene glycol and 0.18 mol KBr in 166g H2O

Carbon w orden

TABLE 13.3 · Molal Boiling-Point-Elevation and Freezing-Point-Depression Constants Normal Boiling Normal Freezing Solvent Point ("C) K. (°C/m) Point (°C) K(°C/m) Water, H2O 100.0 0.51 0.0 1.86 Benzene, CH 80.1 2.53 5.5 Ethanol, C H OH 78.4 1.22 -114.6 1.99 Carbon tetrachloride, CCI 76.8 -22.3 29.8 Chloroform, CHCI 61.2 3.63 -63.5 4.68 5.12 5.02

## Answers

partc per te mass of decane = 18.09 lecane 18 = 0.1268 142 molality = moles 0.127 mass of solvent 0.050 = 2.535 m от; -ку хт To - If = 4.68 * 2.535 1-63.5- Tf = 1.6 = -75.4 freezing point =-75.4 °C oint - 70.4 c parte 0.48 +0.18 molality = molality - 3.976 m 0.166 ATJ = ixkam = 2X1.863-976 T-14.8 freezing point = -14.8. boiling point = 104°c

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