Question

how many valence electrons are present in 0.53g of Na2CO3

Answer 1

configuration of Na = 1s², 2s², 2p^6, 3s¹

so, valence electron in Na = 1

configuration of C = 1s² , 2s² , 2p²

so, valence electrons = 4

configuration of O = 1s², 2s² , 2p⁴

so, valance electrons = 6

hence,

in 1 molecule of Na2CO3 total number of valance electrons = 1 × 2 + 4 + 6 × 3

= 24

hence, in 1 mole of Na2CO3 total number of valance electrons = 24× 6.023 × 10²³

no of mole of Na2CO3 = 0.53/106

= 5 × 10-³

so, total number of valance electrons = 5 × 10-³ × 24 × 6.023 × 10²³

=722.76 × 10^20

= 7.2276 × 10²²