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 answer these questions, we need to apply the concept of apparent volume and apparent mass. The apparent volume and apparent mass are the values obtained during calibration, while the true volume and true mass are the actual values.
The formula to calculate true volume is:
True Volume = Apparent Volume * (Density of Water at 19.0oC / Density of Steel)
The formula to calculate true mass is:
True Mass = Apparent Mass * (Density of Water at 19.0oC / Density of Steel)
Let's calculate the answers step by step:
35. What is the true mass of water delivered in A?
True Mass A = Apparent Mass A * (Density of Water at 19.0oC / Density of Steel)
True Mass A = 10.03 g * (0.9984 g/mL / 7.8 g/mL)
True Mass A = 1.28 g
Therefore, the true mass of water delivered in A is 1.28 g.
36. What is the true mass of water delivered in B?
True Mass B = Apparent Mass B * (Density of Water at 19.0oC / Density of Steel)
True Mass B = 20.09 g * (0.9984 g/mL / 7.8 g/mL)
True Mass B = 2.56 g
Therefore, the true mass of water delivered in B is 2.56 g.
37. What is the true mass of water delivered in C?
True Mass C = Apparent Mass C * (Density of Water at 19.0oC / Density of Steel)
True Mass C = 30.05 g * (0.9984 g/mL / 7.8 g/mL)
True Mass C = 3.85 g
Therefore, the true mass of water delivered in C is 3.85 g.
38. What is the true mass of water delivered in D?
True Mass D = Apparent Mass D * (Density of Water at 19.0oC / Density of Steel)
True Mass D = 40.02 g * (0.9984 g/mL / 7.8 g/mL)
True Mass D = 5.13 g
Therefore, the true mass of water delivered in D is 5.13 g.
39. What is the true volume of water delivered in A?
True Volume A = Apparent Volume A * (Density of Water at 19.0oC / Density of Steel)
True Volume A = 10.05 mL * (0.9984 g/mL / 7.8 g/mL)
True Volume A = 1.29 mL
Therefore, the true volume of water delivered in A is 1.29 mL.
40. What is the true volume of water delivered in B?
True Volume B = Apparent Volume B * (Density of Water at 19.0oC / Density of Steel)
True Volume B = 20.04 mL * (0.9984 g/mL / 7.8 g/mL)
True Volume B = 2.58 mL
Therefore, the true volume of water delivered in B is 2.58 mL.
41. What is the true volume of water delivered in D?
True Volume D = Apparent Volume D * (Density of Water at 19.0oC / Density of Steel)
True Volume D = 40.07 mL * (0.9984 g/mL / 7.8 g/mL)
True Volume D = 5.14 mL
Therefore, the true volume of water delivered in D is 5.14 mL.
42. What is the true volume of water delivered in E?
True Volume E = Apparent Volume E * (Density of Water at 19.0oC / Density of Steel)
True Volume E = 49.98 mL * (0.9984 g/mL / 7.8 g/mL)
True Volume E = 6.41 mL
Therefore, the true volume of water delivered in E is 6.41 mL.
43. What is the correction volume of water delivered in C?
Correction Volume C = True Volume C - Apparent Volume C
Correction Volume C = 3.85 mL - 30.08 mL
Correction Volume C = -26.23 mL
Therefore, the correction volume of water delivered in C is -26.23 mL.
44. What is the true volume of water delivered in D?
We have already calculated the true volume of water delivered in D in question 41, which is 5.14 mL.
Therefore, the true volume of water delivered in D is 5.14 mL.
To determine the true mass and volume of water delivered for each apparent burette reading, we need to account for the density of water at the given temperature (19.0°C). We can use the following formulas:
1. True Mass = Apparent Mass - Correction Mass
2. True Volume = Apparent Volume - Correction Volume
Given information:
- Density of water at 19.0°C = 0.9984 g/mL
- Steel density = 7.8 g/mL
To find the correction mass and correction volume, we need to calculate the difference between the steel and water density.
Correction Mass = Apparent Mass - (Steel density * Apparent Volume)
Now, let's solve each question:
35. What is the true mass of water delivered in A?
True Mass A = Apparent Mass A - Correction Mass A
Correction Mass A = Apparent Mass A - (Steel density * Apparent Volume A)
Substitute the values:
True Mass A = 10.03 g - (7.8 g/mL * 10.05 mL)
Calculate the result: