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15 June, 23:31

Calculate the molarity of each solution. 28.33 g c6h12o6 in 1.28 l of solution

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  1. 16 June, 02:37
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    Molarity is number of moles of solute in 1 L of solution

    number of moles of glucose - 28.33 g / 180 g/mol = 0.1574 mol

    volume of solution is 1.28 L

    since molarity is number of moles in 1 L

    the number of moles in 1.28 L - 0.1574 mol

    therefore number of moles in 1 L - 0.1574 mol / 1.28 L = 0.123 M

    molarity is 0.123 M
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