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At a certain temperature the vapor pressure of pure acetyl bromide (CH_3 COBr) is measured to be 0.75 atm. Suppose a solution is prepared by mixing 51.8 g of acetyl bromide and 123. g of thiophene (C_4H_4S). Calculate the partial pressure of acetyl bromide vapor above this solution. Round your answer to 2 significant digits. Note for advanced students: you may assume the solution is ideal.

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Answer:

0.17 atm

Explanation:

For an  ideal solution we can use Raoult´s law to solve this question:

Pa= XaPºa   (a= acetyl bromide in this case)

where Pa = partial pressure acetyl bromide

          Xa = mole fraction

           Pºa=  vapor pressure of pure acetyl bromide = 0.75 atm

So lets compute the mol fraction of acetyl bromide whice is given by:

Xa = mol of a / ( mol of a + mol of t )          (t stands for thiophene)

We know the masses so find or calculate the molar masses and plug the values:

MW a= 112.95 g/mol  ⇒ mol a = 51.8 g/mol x 1 mol/122.95 g

                                       mol a =  0.42

MW t = 84.14 g/mol    ⇒  mol t = 123 g x  1mol/84.14 g= 1.46 mol

Xa = 0.42/ (0.42 + 1.46) = 0.22

Pa = 0.22 x 0.75 atm = 0.17 atm

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