Sure, let's solve the problem step by step.
Step 1: Determine the weights of the liquids alone.
- The empty bottle weighs [tex]\(25 \, \text{g}\)[/tex].
- The bottle with methylated spirit weighs [tex]\(65 \, \text{g}\)[/tex].
- The bottle with water weighs [tex]\(75 \, \text{g}\)[/tex].
First, find the weight of the methylated spirit alone:
[tex]\[ \text{Weight of methylated spirit} = \text{Weight of bottle with methylated spirit} - \text{Weight of empty bottle} \][/tex]
[tex]\[ \text{Weight of methylated spirit} = 65 \, \text{g} - 25 \, \text{g} = 40 \, \text{g} \][/tex]
Next, find the weight of the water alone:
[tex]\[ \text{Weight of water} = \text{Weight of bottle with water} - \text{Weight of empty bottle} \][/tex]
[tex]\[ \text{Weight of water} = 75 \, \text{g} - 25 \, \text{g} = 50 \, \text{g} \][/tex]
Step 2: Determine the volume of the bottle using the known density of water.
- The density of water is [tex]\(1 \, \text{g/cm}^3\)[/tex].
Since the density of water is known, the volume of the bottle (when filled with water) can be calculated using the weight of the water:
[tex]\[ \text{Volume of water} = \frac{\text{Weight of water}}{\text{Density of water}} \][/tex]
[tex]\[ \text{Volume of water} = \frac{50 \, \text{g}}{1 \, \text{g/cm}^3} = 50 \, \text{cm}^3 \][/tex]
This volume represents the capacity of the bottle, which we can use to find the density of the methylated spirit.
Step 3: Calculate the density of methylated spirit.
- We have the weight of the methylated spirit as [tex]\(40 \, \text{g}\)[/tex].
- We know the volume of the bottle is [tex]\(50 \, \text{cm}^3\)[/tex].
Use the formula for density:
[tex]\[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \][/tex]
[tex]\[ \text{Density of methylated spirit} = \frac{40 \, \text{g}}{50 \, \text{cm}^3} \][/tex]
[tex]\[ \text{Density of methylated spirit} = 0.8 \, \text{g/cm}^3 \][/tex]
Final Answer:
The density of methylated spirits is [tex]\(0.8 \, \text{g/cm}^3\)[/tex].
