How many moles of lithium are present in a sample that contains 2.45x10^87 formula units of li2so4?

The answer is: 8.14·10⁶³ moles of lithium are present.

N(Li₂SO₄) = 2.45·10⁸⁷; number of formula units of lithium sulfate.

n(Li₂SO₄) = N(Li₂SO₄) ÷ Na.

n(Li₂SO₄) = 2.45·10⁸⁷ ÷ 6.022·10²³ 1/mol.

n(Li₂SO₄) = 4.07·10⁶³ mol; amount of lithium sulfate

In one molecule of lithium sulfate, there are two atoms of lithium.

n(Li₂SO₄) : n(Li) = 1 : 2.

n(Li) = 2 · 4.07·10⁶³ mol.

n(Li) = 8.14·10⁶³ mol; amount of lithium atoms.

look at the !

pleaze mark brainliest!

the answer is c

tell me if wrong