Question

concentratia procentuala a solutiei formate prin introducerea unei mase de 6,2g de oxid de sodiu in 100cm cubi de apa

Answer 1

ρH₂O = 1g/cm³
VH₂O = 100cm³

ρ = m/V ⇒ m = ρ×V = 1×100 = 100g H₂O

m = 6,2g Na₂O

MNa₂O = 62g/mol   ,   n = m/M = 6,2/62 = 0,1 moli Na₂O

Na₂O + H₂O ⇒ 2NaOH

1 mol Na₂O.............18g H₂O................2×40g NaOH
0,1 moli Na₂O............xg.............................yg

x = 1,8g H₂O
y = 8g NaOH

md final = 8g 

ms final = 100+8-1,8 = 106,2g

Cf = mdf/msf×100 = 8/106,2×100 = 7,53%