Question

If you were to overshoot the endpoint by 1 drop while you were standardizing the NaOH solution, what would be your % error? Assume the actual volume is 30.00 mL and there are exactly 20 drops in 1.00 ml for the sake of this calculation.

Answer 1

`e(\%)=0.17\%`

Step-by-step explanation:

Volume of a drop:

`V_(drop)=(1 mL)/(20 drop)`

`V_(drop)=0.05 mL`

To estimate the error:

`e(\%)=\frac{V_(real)-V{theorerical}}{V{theoretical}}*100\%`

`e(\%)=((30 mL+0.05mL)-30mL)/(30mL)*100\%`

`e(\%)=0.17\%`