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7 December, 18:39

A solution of phosphoric acid was made by dissolving 10.0 g H3PO4 in 100.0 mL water. The resulting volume was 104 mL. Calculate the density, mole fraction, molarity, and molality of the solution. Assume water has a density of 1.00 g/cm3.

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  1. 7 December, 19:22
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    density is 1.057 g/mL

    mole fraction is 0.0180

    molarity is 0.98 mol/L

    mole fraction is 0.0180

    Explanation:

    Given data

    acid mass = 10 g

    water volume = 100 mL

    total volume = 104 mL

    density = 1 g/cm³

    to find out

    the density, mole fraction, molarity, and molality

    solution

    first we calculate the density that is = total mass g / volume of solution mL

    total mass = mass of H3PO4 + water mass

    so water mass = density x volume

    water mass = 100 ml x 1.0 g/cm3

    water mass = 100 g

    so total mass = 110.00 g

    so that

    density = total mass g / volume of solution mL

    density = 110 / 104 = 1.057 g/mL

    now we calculate no of moles in solvent i. e = mass H2O / mlar mass H2O

    no of moles in solvent = 100/18.015 = 5.55 moles

    and no of moles in solute i. e = mass of H3PO4 / mlar mass in H3PO4

    moles in solute i. e = 10 / 97.994 = 0.102 moles

    so total moles is 5.55 + 0.102 = 5.652 moes

    so now mole fraction = no of moles in solute / total moes

    mole fraction = 5.55 / 5.65

    mole fraction is 0.0180

    now we calculate first

    mole fraction in solute and solvent that is

    mole fraction in solute = no of moles in solute / total moles

    mole fraction in solute = 0.102 / 5.65

    mole fraction in solute is 0.0180

    and mole fraction in solvent that = no of moles in solvent / total moles

    mole fraction in solvent that = 5.55 / 5.65

    mole fraction in solvent that is 0.982

    so molarity = no of moles of solute / volume

    molarity = 0.102/0.104

    molarity is 0.98 mol/L

    and molality is = no of moles of solute / mass

    molality = 0.102 / 100

    molality is 1.02 mol/kg
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