Question

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? Assume that the remainder of the rum has
a density equivalent to water which is 0.997 g/mL.

Answer 1

This problem is providing the density of ethanol as 0.785 g/mL so that the mass of 0.998 L of 75% ethanol in water, with a density of 0.997 g/mL, is required. At the end, the result turns out to be 836 g according to the following:

Density:

In science, density is known as the degree of compactness of a substance and can be easily calculated via:

$\rho =\frac{m}{V}$

However, since this problem is about a mixture comprising ethanol and water, 75% for the former and therefore 25% for the latter, we have to calculate the density of the mixture according to a weighted average:

$\rho_{mix}=0.75*0.785g/mL+0.25*0.997 g/mL=0.838g/mL$

Then, we can solve for the mass in the formula for density to obtain:

$m=\rho * V=0.838g/mL*0.998L*\frac{1000mL}{1L}\\ \\ m=836g$

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