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A microtube is found to have an ungraduated portion that will contain 0.26 g of water. Upon setting up the experimental apparatus as described in the experiment, a microtube is filled to 0.01 ml with water at 4ºC. At a temperature of 60ºC, the volume of head space within the microtube has increased to 0.11 ml. What is the vapour pressure of water at this temperature?

Here is my attempted work. What am I doing incorrect?

n = m x M
= 0.26g x 18.01 g/mol
= 4.6826 mols
PV = nRT
Rearrange to get:
P = nRT/V
= (4.6826 mols)(8.3145 J/K⋅mol)(60ºC)/ 0.11 mL
= 21236.44 atm
Therefore, the vapour pressure of water at this temperature is 21236.44 atm.

  • chemistry -

    You may have other mistakes, also, but 60 degrees C must be substituted into the equation as Kelvin and volume as liters. And if you want the answer to be in atm, then R must be 0.08205.

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