A 3.00 M solution of [tex]C_2H_5OH[/tex] would be prepared by dissolving 0.500 kg of [tex]C_2H_5OH[/tex] in 3.62 kg of water.
Stoichiometric calculation
0.500 kg [tex]C_2H_5OH[/tex] is equivalent to (mole = mass/molar mass):
500g/46.07 = 10.85 moles of [tex]C_2H_5OH[/tex]
3.00 M solution of [tex]C_2H_5OH[/tex] can be prepared by dissolving 10.85 moles in (molarity = moles/volume):
10.85/3 = 3.62 Liters of water
3.62 liters of water weighs (density = mass/volume);
1 x 3.62 = 3.62 kg
Thus, 3.62 kg of water will be required to prepare the solution.
More on stoichiometric calculations can be found here: https://brainly.com/question/27287858
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