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22 March, 02:09

A sulfuric acid solution containing 571.6 g of H2SO4 per liter of solution has a density of 1.329 g/cm3. Calculate (a) the mass percentage, (b) the mole fraction, (c) the molality, (d) the molarity of H2SO4 in this solution.

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  1. 22 March, 03:26
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    a) 43%

    b) 0.122

    c) 7.70 molal

    d) 5.83 M

    Explanation:

    Step 1: Data given

    Mass of H2SO4 = 571.6 grams

    Density of H2SO4 = 1.329 g/cm³ this means 1.329 grams per 1 mL or 1329 grams per 1L

    a) Calculate mass percentage

    Mass % = (571.6 grams / 1329 grams) * 100% = 43%

    b) Calculate mole fraction

    Number of moles H2SO4 = Mass H2SO4 / Molar mass H2SO4

    Moles H2SO4 = 571.6 grams / 98.08 g/mol

    Moles H2SO4 = 5.83 moles

    Moles H2O = (1329 - 571.6) / 18.02 = 42.03 moles

    Total moles = 5.83 + 42.03 = 47.86 moles

    Mole fraction H2SO4 = Moles of solute (H2SO4) / Total moles

    Mole fraction H2SO4 = 5.83 moles / 47.86 moles

    Mol fraction h2SO4 = 0.122

    c) Calculate the molality

    Mass of solvent = 1329 grams - 571.6 grams = 757.4 grams = 0.7574 kg

    Molality of H2SO4 = number of moles H2SO4 / mass of solvent

    Molality H2SO4 = 5.83 moles / 0.7574 kg

    Molality H2SO4 = 7.70 molal

    d) Calculate Molarity

    Molarity H2SO4 = Number of moles H2SO4 / volume

    Molarity H2SO4 = 5.83 moles / 1L = 5.83 M
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