If 15.0 g ethyl alcohol (CH3CH2OH) are dissolved in 500 g water, what is the molality of the solution?
Answer:
molality=moles solute/kg solvent
molar mass of ethyl alcohol= 46g/mol
15/46
0.326
mass of water= 500/1000
0.5kg
molality=0326/0.5
0.652
molality = moles solute / Kg solvent
mass = 500 g = 0.500 Kg
molar mass CH3CH2OH = 46 g/mol
15.0 g / 46 = 0.326 moles
m = 0.326 / 0.500 = 0.652
molality (m) = (moles solute) / (kg solvent)
Moles solute (ethyl alcohol) = 15.0 g / MW
or 15.0 / 46.1 = 0.325 mol
m = 0.354 mol / 0.500 kg = 0.651 m
15.0g ethanol = 11.835ml (density is 0.789g/ml)
Total volume = 500 + 11.835 = 511.835 ml
No. of moles = (V/1000) x concn
15.0g / 46.07 gmol-1 = (511.835/1000) x concn
Therefore concentration = 0.636 M
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