Question

If the actual concentration of the HCl was 0.33M what is your percent error?

Answer 1

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:

=  $\frac{Experimental \ value-Actual \ value}{Actua \ value}\times 100$

On substituting the given values, we get

=  $\frac{0.3532-0.33}{0.33}\times 100$

=  $\frac{0.0225}{0.33}\times 100$

=  $0.06818\times 100$

=  $6.818$