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13 February, 03:22

If a sample of 0.362 m lithium phosphate contains 0.284 g of lithium, what is the volume of the sample?

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Answers (2)
  1. 13 February, 03:57
    0
    From the periodic table:

    mass of lithium = 6.941 grams

    number of moles = mass / molar mass

    number of moles of Li = 0.284 / 6.941 = 0.0409 moles

    Molarity can be calculated using the following rule:

    molarity = number of moles of solute / liters of solution

    volume (in liters) = number of moles of solute / molarity

    volume = 0.0409 / 0.362 = 0.1129 liters
  2. 13 February, 05:13
    0
    Answer;

    = 0.03791 Liters

    Solution and explanation;

    -The molar mass of Li3PO4 is 115.79 g/mol.

    -If lithium is 6.9 g/mol, multiplied by 3 (from Li3PO4), that would be 20.7 g/mol.

    -Now we can use this to find the percentage of Li in Li3PO4, 20.7/115.79 = 0.1788 or 17.88%

    -This 17.88% can tell us the mass of Li3PO4 using the mass which 0.284. Now lets divide 0.284 by 0.1788 = 1.589 g. This now the mass of Li3PO4.

    -Using that, we can now easily find the moles of Li3PO4,

    using the molar mass. So (1.589g) (1mol/115.79g) = 0.013723 moles of Li3PO4.

    -From this, we can now find the volume using molarity (0.362), which is moles per Liter.

    = (0.013723 moles) (1L/0.362mol) = 0.03791 L
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