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CHEMISTRY

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calculate the number of moles of solute present in 124.0g of a solution that is 6.45% glucose by mass.

  • CHEMISTRY -

    6.45% means 6.45 g/100 g soln.
    6.45 x (124.0/100) = 7.998 g in 124.0 g soln.
    moles = grams/molar mass
    moles = 7.998/molar mass glucose. Then round to 3 significant figures.

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