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

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

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