Describe the preparation of 500 mL of 4.75% (w/v) aqueous ethanol (C2H5OH, 46.1 g/mol) .
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1 point
2.38 10-1 g C2 H5 OH
2.38 10+2 g C2 H5 OH
2.38 g C2 H5 OH
23.8 g C2H5OH
To prepare 500 mL of 4.75% (w/v) aqueous ethanol, we need to calculate the amount of ethanol (C2H5OH) needed.
First, calculate 4.75% of 500 mL:
4.75% of 500 mL = 0.0475 * 500 mL = 23.75 mL
Next, we need to convert the volume of ethanol to its mass using its density. The density of ethanol is approximately 0.789 g/mL.
Mass of C2H5OH = Volume * Density
Mass of C2H5OH = 23.75 mL * 0.789 g/mL
Mass of C2H5OH = 18.76 g
Therefore, you would need 18.76 g of ethanol (C2H5OH) to prepare 500 mL of 4.75% (w/v) aqueous ethanol.