Can You Calculate The Number Of Moles Of Na+ Ions In 3.80 L Of 0.220 M Na2SO4?

1 mole of any substance contains 6.023x(10)^23 atoms of that substance.
However,  1 molecule of Na2SO4 breaks down to give 2 Na+ ions.
Thus, 1 mole of Na2SO4 gives 2x6.023x(10)^23 ions of Na+ ions.

1 mole of Na2SO4 is also equal to its molecular weight i.e.
2x23+32+4X16 = 46+32+64 = 142 gms.
Thus,142 gms of Na2SO4 will have 2x6.023x(10)^23 ions of Na+ ions.

Now lets find the gms of Na2SO4 present in 3.8 L of 0.22M Na2SO4:
1 M 1 Lit Na2SO4 = 142 gms
therefore, 0.22 M 1 Lit = 0.22x142 = 31.24gms of Na2SO4.
therfore, 0.22 M 3.8 Lit = 31.24x3.8 = 118.72 gms of Na2SO4.

now, 142 gms of Na2SO4 = 2x6.023x(10)^23 Na+ ions.
Thus, 118.72 gms of Na2SO4 = 2x6.023x(10)^23 x 118.72/142
                                                       = 10.071x(10)^23Na+ ions.

Thus, 3.80 L of 0.22 M Na2SO4 will contain
10.071x(10)^23Na+ ions.