Respuesta :
Answer : The mass percent of Douglasite is 63.75 %
Explanation :
First we have to calculate the moles of [tex]AgNO_3[/tex].
[tex]\text{Moles of }AgNO_3=\text{Molarity of }AgNO_3\times \text{Volume of solution}=0.1mole/L\times 0.03720L=3.72\times 10^{-3}mole[/tex]
As we know that,
Moles of [tex]AgNO_3[/tex] = Moles of AgCl = Moles of [tex]Cl^-[/tex] = [tex]3.72\times 10^{-3}mole[/tex]
The formula of Douglasite is, [tex]2KCl·FeCl_2·2H_2O[/tex]
The molar mass of Douglasite = 311.88 g/mole
Now we have to calculate the moles of [tex]Cl^-[/tex] atoms.
In the formula of Douglasite, there are 4 [tex]Cl^-[/tex] atoms.
[tex]\text{Moles of }Cl^-\text{ atom}=\frac{3.72\times 10^{-3}mole}{4}=9.3\times 10^{-4}mole[/tex]
Now we have to calculate the mass of Douglasite.
[tex]\text{Mass of Douglasite}=\text{Moles of Douglasite}\times \text{Molar mass of Douglasite}[/tex]
[tex]\text{Mass of Douglasite}=(9.3\times 10^{-4}mole)\times (311.88g/mole)=0.29005g=290.05mg[/tex]
conversion used : (1 g = 1000 mg)
Now we have to calculate the mass percent of Douglasite.
[tex]\text{Mass percent of Douglasite}=\frac{\text{Theoretical mass of Douglasite}}{\text{Given mass of Douglasite}}\times 100=\frac{290.05mg}{455.0mg}\times 100=63.75\%[/tex]
Therefore, the mass percent of Douglasite is 63.75 %
