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4 August, 03:31

What is the molarity of a solution prepared from 32.0 grams of methanol (ch3oh, density = 0.792 g/ml) with 125 milliliters of ethanol (ch3ch2oh) ? assume the volumes are additive?

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  1. 4 August, 03:50
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    Molarity = moles of solute / volume of solution

    Mass of methanol = 32.0 grams

    Molar mass of methanol = 32.04 g/mol

    Moles of methanol (solute) = mass / molar mass = 32.0 grams / 32 g/mol

    Moles of methanol (solute) = 1 mole

    Density of methanol = 0.792 g/ml

    Volume of methanol = mass / density = 32.0 g / 0.792 g/ml = 40.4 ml

    Volume of ethanol = 125 ml

    Volume of solution = Volume of methanol (solute) + Volume of ethanol (solvent)

    Volume of solution = 40.4 ml + 125 ml = 165.4 ml = 0.1654 L

    Conversion factor: 1L = 1000ml

    Molarity = 1 / 0.1654 = 6.046 M

    So the molarity of the solution is 6.046 M
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