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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.

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Giangiglia

Answer:

[tex]e(\%)=0.17\%[/tex]

Explanation:

Volume of a drop:

[tex]V_{drop}=\frac{1 mL}{20 drop}[/tex]

[tex]V_{drop}=0.05 mL[/tex]

To estimate the error:

[tex]e(\%)=\frac{V_{real}-V{theorerical}}{V{theoretical}}*100\%[/tex]

[tex]e(\%)=\frac{(30 mL+0.05mL)-30mL}{30mL}*100\%[/tex]

[tex]e(\%)=0.17\%[/tex]

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