# How to Calculate the Mass of a Mixture with Different Densities

## If the density of ethanol is 0.785 g/mL. What should the mass of 0.998 L of rum be in grams if it contains 75% ethanol?

How can we determine the mass of a 0.998 L mixture of 75% ethanol and water with different densities?

## Answer:

The mass of 0.998 L of rum containing 75% ethanol can be calculated by considering the densities of ethanol and water in the mixture.

To begin, we need to calculate the density of the mixture using a weighted average formula. For a mixture containing 75% ethanol and 25% water, the density can be determined as follows:

**Density of Ethanol (ρ _{ethanol}):** 0.785 g/mL

**Density of Water (ρ _{water}):** 0.997 g/mL

Using the weighted average formula:

**Density of Mixture (ρ _{mix}):** 0.75 * 0.785 g/mL + 0.25 * 0.997 g/mL = 0.838 g/mL

Next, we can calculate the mass of the mixture by multiplying the density by the volume:

**Mass (m):** ρ * V = 0.838 g/mL * 0.998 L * 1000 mL/1 L = 836 g

Therefore, the mass of the 0.998 L rum mixture containing 75% ethanol would be 836 grams.

Understanding how to calculate the mass of mixtures with different densities can help in various scientific and practical applications. By considering the proportion of each component and their respective densities, accurate mass calculations can be made.