One student calibrated a 50-mL burette by using the mass of water delivered. The student used an analytical balance which was previously calibrated by STEEL (density = 7.8 g/mL). During calibration, the temperature of water was found to be 19.0oC. The density of water at this temperature is 0.9984 g/mL. The results of the calibration by the student is shown in the table below:
ABCDE Apparent burette reading, mL 10.05 20.04 30.08 40.07 49.98 Weight H2O delivered, g 10.03 20.09 30.05 40.02 49.83
Note: All masses in mentioned in this section are understood to be in grams and all volumes mentioned are understood to be in milliliters.
35. What is the true mass of water delivered in A? A) 10.03
B) 10.04 C) 10.05 D) 10.06 E) 10.07
36. What is the true mass of water delivered in B? A) 20.12
B) 20.13 C) 20.14 D) 20.15 E) 20.16
37. What is the true mass of water delivered in C? A) 30.13
B) 30.11 C) 30.10 D) 30.09 E) 30.08
38. What is the true mass of water delivered in D? A) 40.09
B) 40.08 C) 40.07 D) 40.06 E) 40.05
39. What is the true volume of water delivered in A? A) 10.00
B) 10.02 C) 10.04 D) 10.06 E) 10.08
40. What is the true volume of water delivered in B? A) 20.12
B) 20.14 C) 20.16 D) 20.18 E) 20.20
41. What is the true volume of water delivered in D? A) 40.11
B) 40.12 C) 40.13 D) 40.14 E) 40.15
42. What is the true volume of water delivered in E? A) 50.00
B) 49.98 C) 49.96 D) 49.94 E) 49.92
43. What is the correction volume of water delivered in C? A) 0.04
B) 0.08 C) 0.00 D) -0.04 E) -0.08
44. What is the true volume of water delivered in D? A) 0.11
B) 0.12 C) 0.13 D) 0.14 E) 0.15
To find the true mass of water delivered, we need to calculate the difference between the mass of the water delivered and any air that may be trapped in the burette.
To do this, we need to first calculate the mass of the water delivered in part A. The mass of the water delivered is given as 10.03g.
Next, we need to account for the air trapped in the burette. We know that the density of water is 0.9984 g/mL at 19.0oC.
To find the true mass of water in part A, we can subtract the mass of the air trapped in the burette from the mass of the water delivered.
To calculate the volume of air trapped in the burette, we can use the density of STEEL (density = 7.8 g/mL). Since the density of water is 0.9984 g/mL, the difference in density between water and steel can be used to calculate the volume of air trapped.
The volume of air trapped can be calculated using the formula: Volume of air = Mass of air / Density of air.
The mass of air can be calculated by subtracting the mass of the water from the apparent mass (mass of water + mass of air).
In part A, the apparent mass is 10.05g and the mass of water is 10.03g. Therefore, the mass of air is 10.05g - 10.03g = 0.02g.
We can now calculate the volume of air trapped using the formula: Volume of air = Mass of air / Density of air.
Density of air is given as the difference in density between water and steel, which is 7.8 g/mL - 0.9984 g/mL = 6.8016 g/mL.
Volume of air = 0.02g / 6.8016 g/mL = 0.002936 mL.
Now, we can find the true mass of water in part A by subtracting the volume of air trapped from the mass of the water delivered.
True mass of water in part A = Mass of water - Volume of air trapped = 10.03g - 0.002936 mL * 0.9984 g/mL = 10.03g - 0.00292083g = 10.02707917g.
Therefore, the true mass of water delivered in part A is approximately 10.03g.
To find the true mass of water delivered in parts B, C, D, and E, you can follow the same process using the given data for each part.