The vapor pressure of water is 28.3 mm hg at 28 °c. what mass of water vapor, in mg, would be present in a vapor volume of 600. ml at 28°c? selected b. 16.3 correct answer b. 16.3

Explanation:

The given data is as follows.

       P = 28.3 mm = ,            V = 600 mL = = 0.6 L

      R = 0.082 L atm/mol K,                 T = (28 + 273) K = 301 K

Therefore, according to ideal gas law PV = nRT. Hence, putting the values into this equation calculate the number of moles as follows.

                                    PV = nRT

                  =            

                       n = mol  

As it is known that number of moles equal to mass divided by molar mass. Hence, mass of water vapor present will be calculated as follows. (molar mass of water is 18 g/mol)

               No. of moles =

        mol =

                            mass =

                                      = (approx)

Since, 1 g = 1000 mg. Therefore, will be equal to  

                          = 16.3 mg

Thus, we can conclude that mass of water vapor present is 16.3 mg.

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explanation:

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