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:
*Apparent burette reading, mL: (A)10.05 (B) 20.04 (C) 30.08 (D) 40.07 (E)49.98
*Weight H2O delivered, g: (A)10.03 (B) 20.09 (C)30.05 (D) 40.02 (E) 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?
36. What is the true mass of water delivered in B?
37. What is the true mass of water delivered in C?
38. What is the true mass of water delivered in D?
39. What is the true volume of water delivered in A?
40. What is the true volume of water delivered in B?
41. What is the true volume of water delivered in D?
42. What is the true volume of water delivered in E?
43. What is the correction volume of water delivered in C?
44. What is the true volume of water delivered in D?
To find the true mass of water delivered, we need to use the density of water at 19.0oC. We also need to consider the calibration using the mass of water delivered method.
35. To find the true mass of water delivered in A:
- The apparent burette reading is 10.05 mL.
- The weight of H2O delivered is 10.03 g.
- Since the density of water at 19.0oC is 0.9984 g/mL, the true mass of water delivered can be found using the formula: true mass = apparent volume * density.
- Therefore, the true mass of water delivered in A is 10.05 mL * 0.9984 g/mL = 10.04784 g.
Repeat the same steps for the other questions:
36. True mass of water delivered in B:
- Apparent burette reading = 20.04 mL
- Weight of H2O delivered = 20.09 g
- True mass = 20.04 mL * 0.9984 g/mL = 20.030336 g
37. True mass of water delivered in C:
- Apparent burette reading = 30.08 mL
- Weight of H2O delivered = 30.05 g
- True mass = 30.08 mL * 0.9984 g/mL = 30.062272 g
38. True mass of water delivered in D:
- Apparent burette reading = 40.07 mL
- Weight of H2O delivered = 40.02 g
- True mass = 40.07 mL * 0.9984 g/mL = 40.036128 g
39. True volume of water delivered in A:
- True volume = weight of H2O delivered / density of water at 19.0oC
- True volume = 10.03 g / 0.9984 g/mL = 10.052 mL (approximately)
40. True volume of water delivered in B:
- True volume = weight of H2O delivered / density of water at 19.0oC
- True volume = 20.09 g / 0.9984 g/mL = 20.1207 mL (approximately)
41. True volume of water delivered in D:
- True volume = weight of H2O delivered / density of water at 19.0oC
- True volume = 40.02 g / 0.9984 g/mL = 40.099 mL (approximately)
42. True volume of water delivered in E:
- The true volume can be found by subtracting the apparent burette reading from the volume of water delivered in D.
- True volume = (40.07 mL - 49.98 mL) + true volume of water delivered in D (40.099 mL)
- True volume = 30.161 mL (approximately)
43. Correction volume of water delivered in C:
- Correction volume = true volume - apparent burette reading
- Correction volume = (30.062 mL - 30.08 mL) = -0.018 mL (approximately)
44. True volume of water delivered in D:
- True volume = apparent burette reading + correction volume
- True volume = 40.07 mL + 0.018 mL = 40.088 mL (approximately)