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Question

Water and ethanol, CH3CH2OH(l)\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH}(\mathrm{l}), are miscible, that is, they can be mixed in all proportions. However, when these liquids are mixed, the total volume of the resulting solution is not equal to the sum of the pure liquid volumes, and we say that the volumes are not additive. For example, when 50.0 mL50.0 \mathrm{~mL} of water and 50.0 mL50.0 \mathrm{~mL} of CH3CH2OH(1)\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH}(1), are mixed at 20C20^{\circ} \mathrm{C}, the total volume of the solution is 96.5 mL96.5 \mathrm{~mL}, not 100.0 mL100.0 \mathrm{~mL}. (The volumes are not additive because the interactions and packing of water molecules are slightly different from the interactions and packing of CH3CH2OH\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH} molecules.) Calculate the molarity of CH3CH2OH\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH} in a solution prepared by mixing 50.0 mL50.0 \mathrm{~mL} of water and 50.0 mL50.0 \mathrm{~mL} of CH3CH2OH(l)\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH}(\mathrm{l}) at 20C20^{\circ} \mathrm{C}. At this temperature, the densities of water and ethanol are 0.998210.99821 g/mL\mathrm{g} / \mathrm{mL} and 0.7893 g/mL0.7893 \mathrm{~g} / \mathrm{mL}, respectively.

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Answered 2 years ago
Answered 2 years ago
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To obtain molarity we need two quantities, moles of solute and litres of solution. We must use the molar mass of ethanol to convert from mass to moles and convert the volume from millilitres to litres. Then we can use the definition of molarity to obtain the molarity of the solution.

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