Question

Calculate the % composition of the unknown liquid using your most precise result. It is a mixture of ethanol (D[ETOH] = 0.7890 g/mL) and water. We will assume the temperature was 20.0 °C in the lab and the density of water was D[H2O] = 0.9982 g/mL.

Answer 1

% composition of ethanol = 34.51%

% composition of water  = 65.49%

What is density?

A material's density is defined as its mass per unit volume.

Given data:

The density of ethanol = 0.7890 g/mL

The density of water = 0.9982 g/mL

The density of mixture = 0.926 g/mL

Let the % composition of ethanol = x

Let the % composition of water = 100-x

Now density of the mixture

$\frac{Mass}{Volume}$

$Mass = \frac{percent \;of \;ethanol \;X \;density \;of \;ethanol \;+ \\ \;percent \;of \;water X \;density \;of \;water}{100}$

$0.926 = \frac{x X 0.7890 g/mL \;+ (100-x) X 0.9982 g/mL}{100}$

$x= 34.51$ %

Hence,

% composition of ethanol = 34.51%

% composition of water = 65.49%

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