QQuestionPhysics
QuestionPhysics
When a 10 mL graduated cylinder is filled to the 10 mL mark, the mass of the water was measured to be 10.085 g at 20ºC. If the density of water is taken to be 0.9975 g/mL at 20ºC, what is the percent error for the 10 mL of water?
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Answer
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Step 1:Let's solve this step by step:
Step 2:: Identify the given information
- Accepted density of water at 20ºC: $$0.9975 \mathrm{~g/mL}
- Measured mass: 10.085 \mathrm{~g}
Step 3:: Calculate the expected mass using the accepted density
m_{expected} = V \times \rho = 10 \mathrm{~mL} \times 0.9975 \frac{\mathrm{g}}{\mathrm{mL}} = 9.975 \mathrm{~g}
Step 4:: Calculate the percent error using the formula
\text{Percent Error} = \left|\frac{m_{measured} - m_{expected}}{m_{expected}}\right| \times 100\%
Step 5:: Plug in the values
\text{Percent Error} = \left|\frac{10.085 \mathrm{~g} - 9.975 \mathrm{~g}}{9.975 \mathrm{~g}}\right| \times 100\%
Step 6:: Calculate the percent error
\text{Percent Error} = \left|\frac{0.110 \mathrm{~g}}{9.975 \mathrm{~g}}\right| \times 100\% = 1.10\%
Final Answer
The percent error is 1.10%.
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