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9 January, 15:00

The electrolyte in automobile lead storage batteries is a 3.75 M sulfuric acid solution that has a density of 1.230 g/mL. Calculate the mass percent, molality, and mol fraction of the sulfuric acid.

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  1. 9 January, 15:29
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    Molarity = moles of solute/L of solution

    Molality=moles of solute/kg of solvent

    So, first let's figure out the mass percent: (which would be grams of solute/grams of solution multiplied by 100)

    You have 3.75M sulfuric acid.

    M=Mol Sulfuric Acid/L solution

    To make life easier, let's assume you have 3.75mol sulfuric acid in 1L of solution, which would give us 3.75M

    To find the amount of grams of sulfuric acid that you have, multiply the moles by the molecular weight. So:

    3.75 mol * 68.03 g (molecular weight) : About 255 grams.

    To find out how many grams of solution, convert the amount of solution in liters to grams using the density. So, we assumed we had 1L of solution. Convert 1L to mL (because the density is in g/mL):

    1L * 1000 mL/1L = 1000mL solution

    So density=mass/volume. Plug in the information:

    1.23 g/mL = mass/1000mL

    1.23g/mL * 1000mL=1230 grams solution.

    So, mass percent=

    (255 grams H2SO4/1230 grams solution) * 100%=20.7%

    Molality = moles of solute/kg of solvent

    So, we already assumed that there are 3.75 mol H2SO4 in the solution, so we have the moles solute.

    To determine the amount of solvent:

    -We know there are 255g solute in the solution and 1230 grams in total. And we know, Solute + Solvent = Solution. So plug in, and solve for the solvent:

    255g + Solvent = 1230g

    Solvent=1230g-255g

    We have 3.75 mol H2SO4 and. 975 kg Solvent. Thus, the molality of the solution is:

    mol solute/kg solvent=

    3.75 mol/.975 kg=

    3.85m

    Solvent=975g
  2. 9 January, 17:13
    0
    Mass percent is the amount of solute per amount of solution times 100.

    Mass percent = 3.75 mol / L (1L / 1230 g) (98.08 g / 1 mol) 100 = 27.86%

    Molality is the amount of solute in moles per kilogram solution.

    Molality = 3.75 mol/L (1 L / 1230 g) (1000 g / 1 kg) = 2.84 mol / kg

    Mole fraction is the amount of solute per amount of solution with units of moles.
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