2) Obtain a clean, dry 25-mL graduated cylinder. Weigh the graduated cylinder us

Question

2) Obtain a clean, dry 25-mL graduated cylinder. Weigh the graduated cylinder using a balance Record this value in your lab notebook. 3) Remove the graduated cylinder from the balance and obtain 15-mL of distilled waterla rounding to the nearest 0.1 mL (remember to use the meniscus). Record your vol notebook. 4) ume in your lab Weigh the graduated cylinder and the water using the balance. Record the mass of the graduated cylinder and water in your lab notebook. 5) Obtain the mass of the graduated cylinder with water (a method called weighing by difference). Record this value in your lab notebook. masswater maSScylinder+water maSScylinder mine the density of water using the formula = m/V and record the value in your lab notebook. 7) Perform a percent error calculation (shown below) to determine how far you are from the true value.-, write the observation lexperimental value actual valuel actual value × 100 = percent error 2219 (2.30 Our Water Temperature: Mass of Graduated Cylinder: Mass of Graduated Cylinder with Water: Mass of Water: True Water Density: Experimental Density of Water:04 inenau nens Wa Percent Eror 3041 A percent error of 1% or less would imply you minimized the impact of experimental error in the experiment.

Explanation / Answer

Observations:

Temerature of water = 22oC

Mass of graduated cylinder = 47.77 g

Mass of graduated cylinder along with water = 62.30 g

Volume of water taken = 15 mL

(NOTE: This value should be the actual value taken as read from the lower miniscus from graduated cylinder and not just 15 mL as written in the instructions. It seems either the cylinder is not properly graduated or you have mistakenly taken over 19 mL instead of 15 mL)

Calculations:

Mass of water taken = 62.30 - 47.77 = 19.53 g

Therefor Density of water at 22oC = mass of water / volume of water = 19.53 g/ 15 mL = 1.30 g/mL

True density of water at 22oC (from literature) = 0.9977735 g/mL

Therefor % error = (1.302 - 0.9977735) x 100 /0.9977735 = 30.49 %

(NOTE: If the observed volume is nearly 19.5, then your % error will come down bellow 1%. Please check that.)

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