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:
A B C D E
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

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

35. To find the true mass of water delivered in A, we can use the given density of water at 19.0°C (0.9984 g/mL) and the apparent burette reading (10.05 mL).

True mass = (apparent burette reading) × (density of water)
True mass = 10.05 mL × 0.9984 g/mL = 10.03442 g

The closest value in the options is 10.03 g, so the answer is A) 10.03 g.

36. We do the same for B, using the burette reading of 20.04 mL.

True mass = 20.04 mL × 0.9984 g/mL = 20.011456 g

The closest value from the options is 20.12 g, so the answer is A) 20.12 g.

37. We carry out the same calculation for C, with the burette reading of 30.08 mL.

True mass = 30.08 mL × 0.9984 g/mL = 30.038848 g

The closest value from the options is 30.09 g, so the answer is D) 30.09 g.

38. For D, we use the burette reading of 40.07 mL.

True mass = 40.07 mL × 0.9984 g/mL = 40.066088 g

The closest value from the options is 40.07 g, so the answer is C) 40.07 g.

39. To find the true volume of water delivered in A, we can use the true mass of water and the density of water.

True volume = (true mass) / (density of water)
True volume = 10.03 g / 0.9984 g/mL = 10.044088 mL

The closest value to 10.044088 mL is 10.04 mL, so the answer is C) 10.04 mL.

40. We do the same for B.

True volume = 20.12 g / 0.9984 g/mL = 20.144289 mL

The closest value from the options is 20.14 mL, so the answer is B) 20.14 mL.

41. For D, we use the same calculation.

True volume = 40.07 g / 0.9984 g/mL = 40.110442 mL

The closest value from the options is 40.11 mL, so the answer is A) 40.11 mL.

42. Using the same calculation for E:

True volume = 49.83 g / 0.9984 g/mL = 49.941764 mL

The closest value to 49.941764 mL is 49.94 mL, so the answer is D) 49.94 mL.

43. To find the correction volume of water delivered in C, we can use the true volume and the apparent burette reading.

Correction volume = (true volume) - (apparent burette reading)
Correction volume = 30.038848 mL - 30.08 mL = -0.041152 mL

The closest value from the options is -0.04 mL, so the answer is D) -0.04 mL.

44. For D, we have already calculated the true volume in question 41.

True volume of water delivered in D = 40.11 mL, so the answer is A) 0.11 mL.

To find the true mass of water delivered, we need to correct for the apparent mass of water using the density of water and the temperature correction factor.

Step 1: Calculate the temperature correction factor.
The temperature correction factor (TCF) corrects for the density of water at a temperature other than the reference temperature (usually 20°C). The formula to calculate TCF is as follows:
TCF = (density of water at reference temperature) / (density of water at calibration temperature)

Given values:
density of water at reference temperature = 0.9984 g/mL
density of water at calibration temperature = 0.9984 g/mL (since the temperature during calibration was 19.0°C)

TCF = 0.9984 g/mL / 0.9984 g/mL = 1

Step 2: Calculate the true mass of water delivered.
The true mass of water (TMass) can be calculated using the formula:
TMass = Apparent mass of water / TCF

Given values for the apparent mass of water:
A) 10.03 g
B) 20.09 g
C) 30.05 g
D) 40.02 g
E) 49.83 g

Using the TCF of 1, we can calculate the true mass of water delivered for each apparent mass:

A) TMass A = 10.03 g / 1 = 10.03 g
B) TMass B = 20.09 g / 1 = 20.09 g
C) TMass C = 30.05 g / 1 = 30.05 g
D) TMass D = 40.02 g / 1 = 40.02 g
E) TMass E = 49.83 g / 1 = 49.83 g

Now we can look at the options provided and select the correct answers:

35. The true mass of water delivered in A is 10.03 g. (Option A is correct)
36. The true mass of water delivered in B is 20.09 g. (Option B is correct)
37. The true mass of water delivered in C is 30.05 g. (Option C is correct)
38. The true mass of water delivered in D is 40.02 g. (Option D is correct)
39. Since we have already calculated the true mass of water delivered in A, we can use it to determine the true volume of water delivered in A. The formula to calculate true volume (TVol) is:
TVol = TMass / (Density of water at reference temperature)
TVol A = 10.03 g / 0.9984 g/mL = 10.05 mL (Option C is correct)
40. TVol B = 20.09 g / 0.9984 g/mL = 20.11 mL (Option A is correct)
41. TVol D = 40.02 g / 0.9984 g/mL = 40.07 mL (Option C is correct)
42. True volume is the same as the apparent burette reading. Therefore, the true volume of water delivered in E is 49.98 mL. (Option B is correct)
43. The correction volume is the difference between the true volume and the apparent volume. In this case, the true volume is 30.08 mL and the apparent volume is 30.05 mL. Therefore, the correction volume is 0.03 mL. (Option C is correct)
44. The true volume cannot be determined as there is no apparent mass or volume mentioned for option D.