Answer:
The appropriate solution is "6.818".
Explanation:
The given value is:
Concentration of HCl (Actual value),
= 0.33 M
Concentration of HCl (Experimental value),
= 0.3515 M
Now,
The percentage error (%) will be:
= [tex]\frac{Experimental \ value-Actual \ value}{Actua \ value}\times 100[/tex]
On substituting the given values, we get
= [tex]\frac{0.3532-0.33}{0.33}\times 100[/tex]
= [tex]\frac{0.0225}{0.33}\times 100[/tex]
= [tex]0.06818\times 100[/tex]
= [tex]6.818[/tex]
