Molality of aqueous solution of 8M ethanol having density 1.025g/ml is
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Take a hypothetical sample of 1.000 L of the solution.
(1.000 L) x (8.0 mol/L) x (46.06867 g CH3CH2OH/mol) = 368.55 g CH3CH2OH
((1000 mL) x (1.025 g/mL)) - (368.55 g CH3CH2OH) = 656.45 g H2O
(1.000 L) x (8.0 mol/L) / ((656.45 g H2O) / (1000 g)) = 12 m
(1.000 L) x (8.0 mol/L) x (46.06867 g CH3CH2OH/mol) = 368.55 g CH3CH2OH
((1000 mL) x (1.025 g/mL)) - (368.55 g CH3CH2OH) = 656.45 g H2O
(1.000 L) x (8.0 mol/L) / ((656.45 g H2O) / (1000 g)) = 12 m
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