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
What is the true mass of water delivered in A?
What is the true volume of water delivered in A?
What is the correction volume of water delivered in C?
To find the true mass of water delivered, we need to correct the apparent mass of water delivered by using the density of water at the given temperature.
First, let's focus on row A in the table. The apparent mass of water delivered in A is 10.03 g.
To find the true mass of water delivered in A, we can use the formula:
True mass = Apparent mass / Density of water at given temperature
The density of water at 19.0°C is given as 0.9984 g/mL.
True mass of water delivered in A = 10.03 g / 0.9984 g/mL ≈ 10.06 mL
Therefore, the true mass of water delivered in A is approximately 10.06 g.
To find the true volume of water delivered in A, we can simply use the apparent burette reading, which is 10.05 mL.
Therefore, the true volume of water delivered in A is also 10.05 mL.
Now let's move on to row C in the table. We need to find the correction volume of water delivered in C.
The apparent burette reading in C is 30.08 mL, and the apparent mass of water delivered is 30.05 g.
To find the correction volume, we subtract the true mass of water delivered in C from the apparent mass of water delivered in C, and then divide by the density of water at the given temperature.
Correction mass = Apparent mass - True mass
= 30.05 g - 30.05 g (*True mass equals apparent mass)
= 0 g
Correction volume of water delivered in C = Correction mass / Density of water at given temperature
Since the correction mass is 0 g, the correction volume will also be 0 mL.
Therefore, the correction volume of water delivered in C is 0 mL.