A solution is made by mixing 111.g of benzene C6H6 and 145.g of acetyl bromide CH3COBr...
A solution is made by mixing 111.g of benzene C6H6 and 145.g of
acetyl bromide CH3COBr
Calculate the mole fraction of benzene in this solution. Be sure
your answer has the correct number of significant digits.
Homework Answers
Number of moles of benzene = 111g/78.1118g/mol = 1.42104mol
Number of moles of acetylbromide = 145g/122.9486g/mol = 1.17935mol
Mole fraction of benzene = number of moles of benzene/(total number of moles)
= 1.42104mol/(1.42104+1.17935)mol
= 0.54647
= 0.546 (Answer)
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