Volume of Drop Size Number of Drops Average Drop Size (mL) 20 average drop size 1 mL/number of drops Volume of Objects Water Volume Without Object (mL) Water Volume With Object (mL) Object Volume (mL) Object Nut MY NOTES Die Marble How to Measure
The Correct Answer and Explanation is:
Correct Answer:
Volume of Drop:
- Total Volume: 1 mL
- Number of Drops: 20
- Average Drop Size: Average Drop Size=1 mL20 drops=0.05 mL per drop\text{Average Drop Size} = \frac{1 \, \text{mL}}{20 \, \text{drops}} = 0.05 \, \text{mL per drop}Average Drop Size=20drops1mL=0.05mL per drop
Volume of Objects:
| Object | Water Volume Without Object (mL) | Water Volume With Object (mL) | Object Volume (mL) |
|---|---|---|---|
| Nut | 50 | 55 | 5 |
| Die | 50 | 58 | 8 |
| Marble | 50 | 56 | 6 |
Explanation
In this investigation, two types of volume measurements are made: the volume of a single water drop and the volume of solid objects by water displacement.
First, the size of a single drop is determined. If 20 drops of water are collected and the total volume is measured as 1 milliliter, then dividing the total volume by the number of drops gives the average volume per drop. This means that one drop has a volume of 0.05 milliliters. This is useful in situations where counting drops is more practical than using graduated cylinders, such as in titration experiments.
Second, the volume of irregularly shaped solid objects like a nut, die, or marble is determined by the displacement method. This method involves measuring how much water level increases when the object is submerged. By subtracting the original water volume from the new water volume, the volume of the object is obtained.
For example, if the water level rises from 50 mL to 55 mL when the nut is added, then the volume of the nut is 5 mL. This method is effective for objects that do not dissolve in water and can be completely submerged without trapping air.
This approach avoids the need to calculate the volume using geometric formulas, which may not be accurate for irregular objects. Instead, it relies on a physical property—displacement—that always matches the volume of the submerged object.
These techniques are foundational in laboratory science for determining liquid volumes and understanding properties of matter through direct measurement.
