Answer:
3.50*10^253 L of Helium
Explanation:
I mole of He contains 1 atom
1 mole = 6.022 *10^23 atoms
so 1 mole = 6.022*10^23
(9.42*10^22/ 6.022*10^23) = 9.42*10^22 atoms of He
1.564 *10^45 moles of He = 9.42*10^22 atoms of He
since we know that at S.T.P, 1 mole of a gas contains 22.4L of the gas
1 mole = 22.4 L
1.564*10^45 moles = (1.564*10^45 * 22.4 L)
= 3.50 *10^253 L of He atoms.
So therefore, 9.42*10^22 atoms will contain 3.50*10^253 L of Helium gas.
