The boiling point of water is 100.00 °C at 1 atmosphere. A student dissolves 14.11 grams of sodium nitrate, NaNO3 (85.00 g/mol), in 263.2 grams of water. Use the table of boiling and freezing point constants to answer the questions below. Solvent Formula Kh (°C/m) Ke (°C/m) Water H20 0.512 1.86 Ethanol CH3CH2OH 1.22 1.99 Chloroform CHC13 3.67 Benzene C6H6 2.53 5.12 Diethyl ether CH3CH2OCH2CH3 2.02 The molality of the solution is m. The boiling point of the solution is °C.
The boiling point of water is 100.00 °C at 1 atmosphere. A student dissolves 14.11 grams of sodium nitrate, NaNO3 (85.00 g/mol), in 263.2 grams of water. Use the table of boiling and freezing point constants to answer the questions below. Solvent Formula Kh (°C/m) Ke (°C/m) Water H20 0.512 1.86 Ethanol CH3CH2OH 1.22 1.99 Chloroform CHC13 3.67 Benzene C6H6 2.53 5.12 Diethyl ether CH3CH2OCH2CH3 2.02 The molality of the solution is m. The boiling point of the solution is °C.
Chemistry: Principles and Practice
3rd Edition
ISBN:9780534420123
Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Chapter12: Solutions
Section: Chapter Questions
Problem 12.98QE
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Boiling Point Elevation/Freezing Point Depression
T = m K
where, for freezing point depression:
T = T(pure solvent) - T(solution)
and for boiling point elevation:
T = T(solution) - T(pure solvent)
m | = (# moles solute / Kg solvent) |
Kb | = boiling point elevation constant. |
Kf | = freezing point depression constant. |
Kb and Kf depend only on the SOLVENT. Below are some common values. Use these values for the calculations that follow.
Solvent | Formula | Kb(°C / m) | Kf(°C / m) |
Water | H2O | 0.512 | 1.86 |
Ethanol | CH3CH2OH | 1.22 | 1.99 |
Chloroform | CHCl3 | 3.67 | |
Benzene | C6H6 | 2.53 | 5.12 |
Diethyl ether | CH3CH2OCH2CH3 | 2.02 |
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