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3 July, 04:34

What is the mole fraction, x, of solute and the molality for an aqueous solution that is 16.0 % NaOH by mass

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  1. 3 July, 04:58
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    Step 1) identify the solvent and solute

    solvent = H2O

    solute = NaOH

    Step 2) Assume that there are 100 grams total. Assume that the percent of the compound present is the number of grams present as well

    100 g total

    16g NaOH

    84g H20

    Step 3) convert grams to moles

    84g H2O = 4.64 mol H2O

    19 g NaOH = 0.400 mol NaOH

    Step 4) mole fraction = moles A / (mol A + mol B + ...)

    X = 0.400 / (0.400 + 4.64) = 0.079 (no units)

    Step 5) convert g of solvent to kg

    84 g H2O = 0.084 kg

    Step 6) molality = moles of solute / kg of solvent

    molality = 0.400 mol NaOH / 0.084 kg H2O

    molality = 4.76 m
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