Question

Describe the preparation of 500 mL of 47.5% (w/v) aqueous ethanol (C2H5OH, 46.1 g/mol) .

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1 point
2.38 10-1 g C2 H5 OH
2.3.8 g C2 H5 OH
2.38 10+2 g C2 H5 OH
non of all

Answer 1

[AI-generated answer — Bot GPT 3.5]

The preparation of 500 mL of 47.5% (w/v) aqueous ethanol would involve calculating the amount of ethanol needed based on the desired concentration.

First, calculate the mass of ethanol (C2H5OH) needed:
Percentage concentration = (mass of solute / volume of solution) x 100%
47.5% = (mass of ethanol / 500 mL) x 100%
Mass of ethanol = 47.5/100 x 500 mL = 237.5 g

Next, we need to calculate the amount of ethanol (C2H5OH) in grams required to prepare a 500 mL solution containing 47.5% (w/v) aqueous ethanol.
Density of ethanol = 0.789 g/mL
Volume of ethanol = Mass of ethanol / Density = 237.5 g / 0.789 g/mL = 300.9 mL

To prepare 500 mL of 47.5% (w/v) aqueous ethanol, mix 300.9 mL of ethanol with 199.1 mL of water.