A volume of 10.00 mL of ethanol, CH3CH2OH (density = 0.789 g mL–1 ) is used to make a 100.00 mL aqueous solution with a density of 0.982 g mL–1 .
What is the concentration in mol L–1 of ethanol in the aqueous solution?
Assume that 1 mL of water weighs 1.000 g.
M(CH3CH2OH) = 46.07 g mol–1
Even a hint on where to start would be nice. Thanks
Chemistry Stoichiometry?
2018-02-07 4:14 pm
回答 (1)
2018-02-07 4:25 pm
Mass of CH₃CH₂OH = (10.00 mL) × (0.789 g mL⁻¹) = 7.89 g
Number of moles of CH₃CH₂OH = (7.89 g) / (46.07 g/mol) = 0.171 mol
Volume of the solution = 100.00 mL = (100.00 mL) / (1000 mL/L) = 0.100 L
Concentration of ethanol = (0.171 mol) / (0.100 L) = 1.71 mol/L
(Some data given are unnecessary.)
Number of moles of CH₃CH₂OH = (7.89 g) / (46.07 g/mol) = 0.171 mol
Volume of the solution = 100.00 mL = (100.00 mL) / (1000 mL/L) = 0.100 L
Concentration of ethanol = (0.171 mol) / (0.100 L) = 1.71 mol/L
(Some data given are unnecessary.)
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