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
The physical properties of a solution such as boiling point elevation, freezing point depression, vapor pressure lowering and osmotic pressure are together called as colligative properties which depend on the number of solute particles or ions and independent of identity of ions.
Difference in boiling point before an addition of solute and after the addition of solute is directly proportional to molality of a solute.
For the two solutions with the same boiling point, the molal concentration can be calculated by equating.
Elevation of boiling point:
Normal boiling point is a temperature at which liquid state changes to gaseous state at one atmospheric pressure. Elevation of boiling point is one of the colligative properties that work upon the addition of solute molecules to pure solvent.
Mathematically,
Given that,
The boiling point of benzene solution =
The boiling point of carbon tetrachloride =
The molal boiling point elevation constant of benzene
The molal boiling point elevation constant of carbon tetrachloride
Since, the two solutions have the same boiling point.
The boiling point is calculated below:
Ans:
The molal concentration, m (or b), and boiling point, Tb is determined below:
Imagine two solutions with the same concentration and the same boiling point, but one has benzene as the solvent and th...
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
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.
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...
Assuming 100% dissociation, calculate the freezing point and boiling point of 2.11 m Na2SO4(aq). 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
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...
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...
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...
What is the boiling point of a 0.66 m solution of a nonvolatile solute in benzene? Constants: Kb = 2.53 oCm-1 Tb= 80.1 oC
4. Kbp for benzene is 2.53°C/molal, and its normal boiling point is 80.1°C. What is the boiling point of a solution of 11.6 g of CsH12N20, which is not ionic and has a molar mass of 96.17 g/mol, in 80.0 g of benzene? DTG = kbmit A. 64.3°C B. 76.9°C 80.1= 2.53 C. 83.3°C D. 95.9°C 1Kbp=2.53 T6=80.1°C To = ? 11. 6 graco H₂ N20 0.0118 kg 96.179 1810 398 25309 (80.1) 896. do