Calculate the molality of 1.5g of acetic acid----CH3COOH in 80g of benzene.
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The molar mass of ethanoic acid is 2(12)+2(16)+4(1)=60 g/mol. The mass of ethanoic acid is divided with molar mass of ethanoic acid to obtain number of moles of ethanoic acid.
Number of moles of ethanoic acid =
60g/mol
2.5g
=0.04167mol
The mass of benzene is 75 grams or
1000
75
=0.075 kg.
The molality of ethanoic acid in benzene is the ratio of the number of moles of ethanoic acid to the mass of benzene (in kg).
It is
0.075kg
0.04167mol
=0.556m
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