calculate molality of 2.5g of ethanoic acid in 75g of benzene
sanya55:
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Answered by
271
CALCULATIONS
molality = mol / kg
Weight of CH₃COOH=2.5 gm
molar weight of CH₃COOH= 12+3+12+16 x2 +1=60
Weight of benzene= 75 gm =0.075 kg
Molar weight of C6H6= 12 x6 +1 x 6 =78
Molality =weight of solute/ (molar weight of solute x weight of solvent in kg)
m=2.5/60 x 0.075
m =0.56 mole /kg
molality = mol / kg
Weight of CH₃COOH=2.5 gm
molar weight of CH₃COOH= 12+3+12+16 x2 +1=60
Weight of benzene= 75 gm =0.075 kg
Molar weight of C6H6= 12 x6 +1 x 6 =78
Molality =weight of solute/ (molar weight of solute x weight of solvent in kg)
m=2.5/60 x 0.075
m =0.56 mole /kg
Answered by
32
Answer:
Weight of ch3cooh =2.5g
Molar weight of ch3cooh= 12+3+12+16×2+1=60
Molar weight of benzene =78
Molality = 2.5/60×0.075
=0.56mole/kg
Explanation:
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