Question

A particular small laboratory cork weighs \(1.63 \mathrm{g}\) whereas a rubber lab stopper of the same size weighs 4.31 g. How many corks would there be in 500. g of such corks? How many rubber stoppers would there be in \(500 .\) g of similar stoppers? How many grams of rubber stoppers would be needed to contain the same number of stoppers as there are corks in \(1.00 \mathrm{kg}\) of corks?

Short answer

In 500g of corks, there are 307 corks and in 500g of rubber stoppers, there are 116 rubber stoppers. To have the same number of rubber stoppers as there are corks in 1.00kg of corks, 2.64634kg of rubber stoppers would be required.

Step-by-step solution

Find the number of corks in 500g

To find the number of corks in 500g, divide the total weight (500g) by the weight of one cork (1.63g): Number of corks = Total weight / Weight of one cork Note that the units will cancel each other out as we are dividing grams by grams.

Calculate the number of corks in 500g

Using the formula from Step 1: Number of corks = 500g / 1.63g Number of corks = 306.748 Since we cannot have a fraction of a cork, we round the number to the nearest whole number: Number of corks = 307 There are 307 corks in 500g of corks.

Find the number of rubber stoppers in 500g

To find the number of rubber stoppers in 500g, divide the total weight (500g) by the weight of one rubber stopper (4.31g): Number of rubber stoppers = Total weight / Weight of one rubber stopper Note that the units will cancel each other out as we are dividing grams by grams.

Calculate the number of rubber stoppers in 500g

Using the formula from Step 3: Number of rubber stoppers = 500g / 4.31g Number of rubber stoppers = 116.001 Since we cannot have a fraction of a rubber stopper, we round the number to the nearest whole number: Number of rubber stoppers = 116 There are 116 rubber stoppers in 500g of rubber stoppers.

Find the number of corks in 1.00kg of corks

First, convert the weight from 1.00kg to g: 1.00kg = 1000g Now use the formula from Step 1 to find the number of corks: Number of corks = Total weight / Weight of one cork

Calculate the number of corks in 1.00kg of corks

Using the formula from Step 5: Number of corks = 1000g / 1.63g Number of corks = 613.497 Since we cannot have a fraction of a cork, we round the number to the nearest whole number: Number of corks = 614 There are 614 corks in 1.00kg of corks.

Find the weight of rubber stoppers to have the same number of stoppers as corks in 1.00kg of corks

To find the weight of rubber stoppers needed to have the same number of stoppers as corks in 1.00kg of corks, multiply the number of corks (614) by the weight of one rubber stopper (4.31g): Weight of rubber stoppers = Number of stoppers * Weight of one rubber stopper

Calculate the weight of rubber stoppers needed

Using the formula from Step 7: Weight of rubber stoppers = 614 * 4.31g Weight of rubber stoppers = 2646.34g To express the result in kg, divide by 1000: Weight of rubber stoppers = 2.64634kg 2.64634kg of rubber stoppers would be needed to contain the same number of stoppers as there are corks in 1.00kg of corks.

Key concepts

mass and volume conversion

Understanding mass and volume conversion is key in chemistry, allowing you to switch between different units and work with measurable quantities effectively.
To convert mass from kilograms (kg) to grams (g), which is common in laboratory settings, simply multiply the mass value by 1000, because there are 1000 grams in a kilogram.
This conversion is essential in experiments where precise measurements are crucial. In the original exercise, the corks in 1.00 kg were converted to 1000 g to facilitate calculation. This ensures accuracy when calculating the number of items based on their individual weights.
Remember, keeping track of the units during conversion helps in avoiding errors. Use unit conversion factor tables or calculators to aid in swift conversions without confusion.

weight comparison

Weight comparison involves understanding and contrasting weights of different objects with similar or varying sizes and densities.

• In the given problem, the cork and rubber stopper of the same size have different weights (1.63 g and 4.31 g, respectively).


• Such differences are due to the density of the materials, with rubber generally being denser than cork.

Comparing weights allows us to deduce certain properties; for example, heavier materials at the same volume might hint at higher densities.
Weight comparison is fundamental when choosing materials for specific applications, influencing decisions based on weight constraints or capacity requirements.

material density

Material density is a measure of how much mass is contained in a given volume.
It plays a critical role in determining how a material behaves in different conditions and applications.

• It is often expressed in units of grams per cubic centimeter (g/cm³).
• Density helps in comparing materials by showing how compacted their mass is per unit volume.

In our example, the heavier rubber stopper indicates it has a higher density compared to the cork.
Understanding material density helps in predicting how substances will interact with each other and their surroundings.
Knowing the density aids in calculations involving buoyancy, purity analysis, and component selection for manufacturing.

unit conversion in chemistry

Unit conversion in chemistry is essential because it helps in translating different measurement units, enabling clear communication and comparison across scientific communities. In many chemistry problems, such as the exercise given, one often deals with mass (grams, kilograms) and needs to switch between these units depending on the context or requirement.
Here are some key points about unit conversion:

• Always write down the conversion factor, like \(1 ext{ kg} = 1000 ext{ g} \).
• Use dimensional analysis to ensure that units cancel out properly during conversion, minimizing errors.

Proper unit conversion ensures the integrity and accuracy of your calculations.
It helps maintain consistency across various scenarios and ensures laboratory and theoretical results align.