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Paha777 [63]
2 years ago
11

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.
Chemistry
1 answer:
Rufina [12.5K]2 years ago
8 0

% composition of ethanol = 34.51%

% composition of water  = 65.49%

<h3>What is density?</h3>

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%

Learn more about the density here:

brainly.com/question/952755

#SPJ1

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