A solution is prepared by dissolving 5.00 g of glycerin () in 201 g of ethanol The freezing point of the solution is ________°C. The freezing point of pure ethanol is -114.6 °C at 1 atm. The molal-freezing-point-depression constant () for ethanol is 1.99 °C/m. The molar masses of glycerin and of ethanol are 92.1 g/mol and 46.1 g/mol, respectively.
-115.1 |
-119.4 |
0.537 |
-109.8 |
-114.1 |
A solution is prepared by dissolving 5.00 g of glycerin () in 201 g of ethanol The...
A solution is prepared by dissolving 2.00 g of glycerin (C3H8O3) in 201 g of ethanol (C2H5OH). The freezing point of the solution is _______ degrees celsius. The freezing point of pure ethanol is -114.6 degrees celsius at 1atm. The molal-freezing point depression constant (Kf) for ethanol is 1.99 degrees celsius/m. The molar masses of glycerin and of ethanol are 92.1 g/mol and 46.1 g/mol, respectively.
2. A solution is prepared by dissolving 7.00 g of glycerin (C3H8O3) in 201 g of ethanol (C2H5OH). The freezing point of the solution is __________°C. The freezing point of pure ethanol is -114.6 °C at 1 atm. The molal-freezing-point-depression constant (Kf) for ethanol is 1.88 °C/m. The molar masses of glycerin and of ethanol are 92.1 g/mol and 46.1 g/mol, respectively.
9. 65) A soltion is prepared by dissolving 7.00 g of gycerin (C3Hg03) in 201 g of ethanol (C2HsOH). The freezing point of the solution isoC freezing-point-depression constant (Kp for ethanol is 1.99°C/m. The molar °C. The feezing point of pure ethanol is -114.6°C at 1 atm. The molal- masses of glycerin and of ethanol are 92.1 g/mol and 46.1 g/mol respectively.
What is the freezing point (°C) of a solution prepared by dissolving 11.3 g of in 115 g of water? [Use these Molar Masses: Ca = 40, N = 14, O = 16. Also, the molal freezing point depression constant for water is 1.86 °C/m.]
Benzaldehyde ( MM = 106.18 g/mol) also known as oil of almonds is used in the manufracture of dyes and perfume and in flavorings. what would be the freezing point of a solution prepared by dissolving 106.345 g of benzaldehyde in 1.687 kg of ethanol? kf=1.99, freezing point of pure ethanol = -117.3 degree Celsius a) moles of benzaldehyde b) kg of ethanol c) molality d) freezing point depression : delta Tf= kf×m e) freezing point of the solution
Given the following information. Determine the freezing points of a solution which contains 76.5 g of a sodium carbonate (Na2CO3) in 500.0 g of ethanol (C2H6O). Kf for ethanol 1.99 °C / m Normal freezing point for ethanol -114.6 °C
a) Formaldehyde (CH2=O) is the simplest aldehyde and is miscible in ethanol. i) Would the following intermolecular interactions be present or absent for an ethanol-formaldehyde mixture: hydrogen bonding, London dispersion, ion-ion, and dipole-induced dipole. [8 marks] ii) 66 g of formaldehyde is dissolved in 800 g of ethanol. Given that the density of ethanol 789 g L-1, calculate the molarity, molality, and mole fraction for this solution. [20 marks] iii) Ethanol has a freezing point depression coefficient Kfp = -1.99...
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 0.592 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...