Imagine two solutions with the same concentration and the same boiling point, but one has benzene as the solvent and the other has carbon tetrachloride as the solvent. Determine that molal concentration, m (or b), and boiling point, Tb.
Imagine two solutions with the same concentration and the same boiling point, but one has benzene...
Imagine two solutions with the same concentration and the same boiling point, but one has benzene as the solvent and the other has carbon tetrachloride as the solvent. Determine that molal concentration, m (or b), and boiling point, Tb. benzene boiling point=80.1 Kb=2.53 carbon tetrachloride boiling point=76.8 Kb=5.03
Imagine two solutions with the same concentration and the Normal boiling point к, (С/m) Solvent same boiling point, but one has benzene as the solvent and (C) the other has carbon tetrachloride as the solvent. Determine benzene 80.1 2.53 the molal concentration, m (or b), and boiling point, T. carbon tetrachloride 76.8 5.03 Т, т m °C
a Calculate the freezing point and the boiling point of each of the following aqueous solutions. (Assume complete dissociation. Assume that water freezes at 0.00°C and boils at 100.000ºC.) 0.047 m MgCl2 Kr=-1.86 °C/molal Kb =0.51 °C/molal Tf= °C Tb = °C Submit b Calculate the freezing point and the boiling point of each of the following aqueous solutions. (Assume complete dissociation. Assume that water freezes at 0.00°C and boils at 100.000°C.) 0.047 m FeCl3 Kf=-1.86 °C/molal Kb =0.51 °C/molal...
Molal Boiling-Point-Elevation and Freezing-Point-Depression Solvent Normal Boiling Point (∘C) Kb (∘C/m) Normal Freezing Point (∘C) Kf (∘C/m) Water, H2O 100.0 0.51 0.0 1.86 Benzene, C6H6 80.1 2.53 5.5 5.12 Ethanol, C2H5OH 78.4 1.22 -114.6 1.99 Carbon tetrachloride, CCl4 76.8 5.02 -22.3 29.8 Chloroform, CHCl3 61.2 3.63 -63.5 4.68 Part E freezing point of 2.02 g KBr and 4.84 g glucose (C6H12O6) in 187 g of water Part F boiling point of 2.02 g KBr and 4.84 g glucose (C6H12O6) in...
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...
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...
A solution is made by dissolving 0.592 mol of nonelectrolyte solute in 767 g of benzene. Calculate the freezing point, Te, and boiling point, Tb, of the solution. Constants can be found in the table of colligative constants. T = Colligative Constants Constants for freezing-point depression and boiling-point elevation calculations at 1 atm: Solvent Formula Kf value* Normal freezing Kb value Normal boiling (°C/m) point (°C) (°C/m) point (°C) water H20 1.86 0.00 0.512 100.00 benzene 5.12 5.49 2.53 80.1...
A solution is made by dissolving 0.585 mol of nonelectrolyte solute in 877 g of benzene. Calculate the freezing point, Tf, and boiling point, Tb, of the solution. Constants can be found in the table of colligative constants. Constants for freezing-point depression and boiling-point elevation calculations at 1 atm: 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. Benzene and toluene form nearly ideal solutions. The boiling point of pure benzene is 80.1°C. Calculate the chemical potential of benzene relative to that of pure benzene when Xbenzene = 0.30 at its boiling point. If the activity coefficient of benzene in this solution were actually 0.93 rather than 1.00, what would be its vapor pressure?
Benzene and toluene form nearly ideal solutions. The boiling point of pure benzene is 80.1°C. Calculate the chemical potential of benzene relative to that of pure benzene when χbenzene = 0.30 at its boiling point. If the activity coefficient of benzene in this solution were actually 0.93 rather than 1.00, what would be its vapor pressure?