Problem 50
Question
To determine the volume of an irregularly shaped glass vessel, the vessel is weighed empty \((121.3 \mathrm{g})\) and when filled with carbon tetrachloride (283.2 g). What is the volume capacity of the vessel, in milliliters, given that the density of carbon tetrachloride is \(1.59 \mathrm{g} / \mathrm{mL} ?\)
Step-by-Step Solution
Verified Answer
The volume capacity of the vessel is 101.89 mL.
1Step 1: Determine the mass of carbon tetrachloride
This is done by subtracting the weight of the empty vessel from the weight of the vessel when filled with carbon tetrachloride. So, \(mass_{carbon\ tetrachloride} = weight_{filled\ vessel} - weight_{empty\ vessel}\) = \(283.2 g - 121.3 g\) = \(161.9 g\)
2Step 2: Apply density formula to find the volume
We know that density = mass/volume. By rearranging, we get volume = mass/density. Hence, \(volume_{carbon\ tetrachloride} = mass_{carbon\ tetrachloride} / density_{carbon\ tetrachloride}\) = \(161.9g / 1.59 g/mL\) = \(101.885 mL\)
3Step 3: Round to appropriate significant figures
Considering the provided values in the problem, the measurement with the least number of decimal places is of two places, hence we need to round our answer to two decimal places. Therefore, \( volume_{final} = 101.89 mL\)
Key Concepts
Volume MeasurementSignificant FiguresMass and Weight
Volume Measurement
Understanding how to measure volume, especially for irregularly shaped objects, is key to solving many chemical and physical problems. Volume can be thought of as the amount of space an object occupies. For regular shapes like cubes or spheres, we can use mathematical formulas to calculate their volume. However, for irregularly shaped objects, we often use a method involving the displacement of a liquid or rely on weight differences.
In this exercise, the vessel's volume was determined by filling it with a liquid and measuring the mass difference. By knowing the mass of the carbon tetrachloride added to the vessel and its known density, we can find the volume. This is possible because density is defined as mass per unit volume, and rearranging the density formula allows us to derive the volume.
The formula used in this context is:
In this exercise, the vessel's volume was determined by filling it with a liquid and measuring the mass difference. By knowing the mass of the carbon tetrachloride added to the vessel and its known density, we can find the volume. This is possible because density is defined as mass per unit volume, and rearranging the density formula allows us to derive the volume.
The formula used in this context is:
- Volume = Mass / Density
Significant Figures
Significant figures are the digits in a number that contribute to its precision, affecting how we report measurements and calculate results in scientific contexts. When performing calculations, especially in experiments, it's important to express results with the correct number of significant figures. This ensures that the precision of your measurement is accurately communicated, neither overestimating nor underestimating it.
In this particular exercise, we dealt with subtraction and division using values that have varying degrees of precision. When you subtract values, like the two weights in the problem, the result should reflect the least precise measurement. In the division for calculating volume, significant figures from the processed mass carrying the least decimal place define the precision of your final result. Since the reported mass and density both have values to one decimal place, the calculated volume should reflect two decimal places. Thus, the final result is rounded appropriately to 101.89 mL.
In this particular exercise, we dealt with subtraction and division using values that have varying degrees of precision. When you subtract values, like the two weights in the problem, the result should reflect the least precise measurement. In the division for calculating volume, significant figures from the processed mass carrying the least decimal place define the precision of your final result. Since the reported mass and density both have values to one decimal place, the calculated volume should reflect two decimal places. Thus, the final result is rounded appropriately to 101.89 mL.
Mass and Weight
While often used interchangeably in day-to-day language, mass and weight hold distinct meanings in scientific terms. Mass refers to the amount of matter an object contains and remains constant regardless of location. It is usually measured in grams or kilograms. Weight, on the other hand, depends on the gravitational pull exerted on that mass and can vary depending on where you are in the universe.
In the context of this exercise, when we refer to weighing the vessel empty and filled, we are indirectly measuring mass even though we often say weight. The difference in these measurements allows us to determine the mass of the carbon tetrachloride alone. Knowing this mass is vital because it links directly to calculating volume when used in conjunction with the known density. Weight, in scientific calculations, ensures accuracy in determining how much a particular substance is being considered, directly influencing subsequent volume and density computations.
In the context of this exercise, when we refer to weighing the vessel empty and filled, we are indirectly measuring mass even though we often say weight. The difference in these measurements allows us to determine the mass of the carbon tetrachloride alone. Knowing this mass is vital because it links directly to calculating volume when used in conjunction with the known density. Weight, in scientific calculations, ensures accuracy in determining how much a particular substance is being considered, directly influencing subsequent volume and density computations.
Other exercises in this chapter
Problem 47
A 2.18 L sample of butyric acid, a substance present in rancid butter, has a mass of 2088 g. What is the density of butyric acid in grams per milliliter?
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A solution consisting of \(8.50 \%\) acetone and \(91.5 \%\) water by mass has a density of \(0.9867 \mathrm{g} / \mathrm{mL} .\) What mass of acetone, in kilog
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A solution contains \(10.05 \%\) sucrose (cane sugar) by mass. What mass of the solution, in grams, is needed for an application that requires \(1.00 \mathrm{kg
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