To find the density of the mixture in SI units, we will follow a structured approach.
1. Convert the volumes from cubic centimeters to cubic meters:
Given:
- Volume of alcohol = [tex]\( 400 \, \text{cm}^3 \)[/tex]
- Volume of water = [tex]\( 600 \, \text{cm}^3 \)[/tex]
- Conversion factor = [tex]\( 1 \, \text{cm}^3 = 1 \times 10^{-6} \, \text{m}^3 \)[/tex]
So,
- Volume of alcohol in [tex]\( \text{m}^3 \)[/tex] = [tex]\( 400 \, \text{cm}^3 \times 1 \times 10^{-6} \, \text{m}^3/\text{cm}^3 = 0.0004 \, \text{m}^3 \)[/tex]
- Volume of water in [tex]\( \text{m}^3 \)[/tex] = [tex]\( 600 \, \text{cm}^3 \times 1 \times 10^{-6} \, \text{m}^3/\text{cm}^3 = 0.0006 \, \text{m}^3 \)[/tex]
2. Calculate the masses of alcohol and water:
Given:
- Density of alcohol = [tex]\( 800 \, \text{kg}/\text{m}^3 \)[/tex]
- Density of water = [tex]\( 1000 \, \text{kg}/\text{m}^3 \)[/tex]
Using the formula [tex]\( \text{mass} = \text{density} \times \text{volume} \)[/tex],
- Mass of alcohol = [tex]\( 800 \, \text{kg}/\text{m}^3 \times 0.0004 \, \text{m}^3 = 0.32 \, \text{kg} \)[/tex]
- Mass of water = [tex]\( 1000 \, \text{kg}/\text{m}^3 \times 0.0006 \, \text{m}^3 = 0.6 \, \text{kg} \)[/tex]
3. Calculate the total mass and total volume of the mixture:
- Total mass of the mixture = Mass of alcohol + Mass of water
= [tex]\( 0.32 \, \text{kg} + 0.6 \, \text{kg} = 0.92 \, \text{kg} \)[/tex]
- Total volume of the mixture = Volume of alcohol + Volume of water
= [tex]\( 0.0004 \, \text{m}^3 + 0.0006 \, \text{m}^3 = 0.001 \, \text{m}^3 \)[/tex]
4. Calculate the density of the mixture:
Using the formula [tex]\( \text{density} = \frac{\text{total mass}}{\text{total volume}} \)[/tex],
- Density of the mixture = [tex]\( \frac{0.92 \, \text{kg}}{0.001 \, \text{m}^3} = 920 \, \text{kg}/\text{m}^3 \)[/tex]
Thus, the density of the mixture is [tex]\( 920 \, \text{kg}/\text{m}^3 \)[/tex].
