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What volume of water, with a density of 1.00 g/mL, must be added to 25.0 g of CH₃CH₂OH to form a solution that is 0.100 molality?

Select one:
A. 5.43 L
B. 250 mL
C. 2.50 L
D. 543 mL
E. 4.00 L

Sagot :

GinnyAnswer
To solve this problem, let's go through the steps step-by-step:

1. Determine the molar mass of ethanol (CH₃CH₂OH):
- Carbon (C) has an atomic mass of 12.01 g/mol.
- Hydrogen (H) has an atomic mass of 1.008 g/mol.
- Oxygen (O) has an atomic mass of 16.00 g/mol.
- The molecular formula for ethanol is C₂H₆O.

Calculating the molar mass:
- [tex]\(2 \times 12.01\)[/tex] for the carbon atoms.
- [tex]\(6 \times 1.008\)[/tex] for the hydrogen atoms.
- [tex]\(1 \times 16.00\)[/tex] for the oxygen atom.

Summing these values:
[tex]\[ 2 \times 12.01 + 6 \times 1.008 + 16.00 = 24.02 + 6.048 + 16.00 = 46.068 \text{ g/mol} \][/tex]

2. Calculate the number of moles of ethanol:
- Given mass of ethanol: 25.0 g.
- Using the molar mass of ethanol: 46.068 g/mol.

The number of moles of ethanol is:
[tex]\[ \text{moles of ethanol} = \frac{\text{mass of ethanol}}{\text{molar mass of ethanol}} = \frac{25.0 \text{ g}}{46.068 \text{ g/mol}} \approx 0.54268 \text{ mol} \][/tex]

3. Use the definition of molality (mol/kg):
- Molality (m) is defined as moles of solute per kilogram of solvent.
- Given molality: 0.100 mol/kg.

Rearranging the formula to find the mass of the solvent (water):
[tex]\[ \text{mass of water (kg)} = \frac{\text{moles of ethanol}}{\text{molality}} = \frac{0.54268 \text{ mol}}{0.100 \text{ mol/kg}} \approx 5.427 \text{ kg} \][/tex]

4. Convert the mass of water from kg to grams:
- Since [tex]\(1 \text{ kg} = 1000 \text{ g}\)[/tex]:
[tex]\[ \text{mass of water (g)} = 5.427 \text{ kg} \times 1000 \text{ g/kg} = 5427.76 \text{ g} \][/tex]

5. Determine the volume of water:
- The density of water is given as 1.00 g/mL.
- Since density is mass per unit volume, for water, [tex]\(1 \text{ g} = 1 \text{ mL}\)[/tex].

Therefore, the volume of water is:
[tex]\[ \text{volume of water} = 5427.76 \text{ g} = 5427.76 \text{ mL} \][/tex]

6. Convert the volume to liters if needed:
- Since [tex]\(1 \text{ L} = 1000 \text{ mL}\)[/tex]:
[tex]\[ \text{volume of water (L)} = \frac{5427.76 \text{ mL}}{1000 \text{ mL/L}} \approx 5.43 \text{ L} \][/tex]

Therefore, the volume of water that must be added is approximately 5.43 liters.

The correct answer is:
a. 5.43 L
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