Answer:
The chemical formula of this residue: [tex]Na_2CrO_4.10H_2O[/tex]
Explanation:
Mass of residue = 19.12
Let the formula of the residue be [tex]Na_2CrO_4.xH_2O[/tex]
Moles of residue: n
[tex]n=\frac{19.12}{162 g/mol+x\times 18 g/mol}[/tex]
Moles of sodium chromate = n'
Molarity of sodium chromate = 0.1035 M
Volume of sodium chromate solution = 540.0 mL = 0.540 L
1 mL = 0.001 L
[tex]Concentration=\frac{moles}{Volume(L)}[/tex]
[tex]0.1035 M=\frac{n}{0.540 L}[/tex]
[tex]n'=0.1035 M\times 0.540 L=0.05589 mol[/tex]
[tex]Na_2CrO_4+xH_2O\rightarrow Na_2CrO_4.xH_2O[/tex]
According to reaction,1 mole of [tex]Na_2CrO_4[/tex] gives 1 mole of[tex]Na_2CrO_4.xH_2O[/tex]
So, n = n'
[tex]\frac{19.12}{162 g/mol+x\times 18 g/mol}=0.05589 mol[/tex]
x = 10
The chemical formula of this residue: [tex]Na_2CrO_4.10H_2O[/tex]
