Tetrahydrofuran (commonly referred to as THF) is an organic solvent with a density of 0.889 g/mL. A solution is prepared by dissolving 11.8 g of dichlorobenzene (C6H4Cl2, MW = 147.00 g/mol) in 100.0 mL of THF. What is the molality (m) of dichlorbenzene in the THF-based solution? D A) 0.903 B) 0.0889 C) 0.0803 D) 1.11 No
The Correct Answer and Explanation is :
Problem Breakdown and Solution
Molality ((m)) is defined as the number of moles of solute per kilogram of solvent:
[
m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}
]
Here are the steps to calculate the molality:
1. Moles of Dichlorobenzene:
We are given:
- Mass of dichlorobenzene ((C_6H_4Cl_2)) = 11.8 g
- Molar mass of (C_6H_4Cl_2) = 147.00 g/mol
Calculate the moles of (C_6H_4Cl_2):
[
\text{moles of solute} = \frac{\text{mass}}{\text{molar mass}} = \frac{11.8}{147.00} = 0.0803 \, \text{mol}
]
2. Mass of Solvent (THF):
We are given:
- Volume of THF = 100.0 mL
- Density of THF = 0.889 g/mL
Calculate the mass of THF:
[
\text{mass of solvent} = \text{density} \times \text{volume} = 0.889 \, \text{g/mL} \times 100.0 \, \text{mL} = 88.9 \, \text{g}
]
Convert the mass to kilograms:
[
\text{mass of solvent in kg} = \frac{88.9}{1000} = 0.0889 \, \text{kg}
]
3. Calculate Molality:
Substitute the values into the molality formula:
[
m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.0803}{0.0889} = 0.903 \, \text{mol/kg}
]
Final Answer:
The molality of dichlorobenzene in the THF-based solution is:
[
\boxed{0.903 \, \text{mol/kg}} \, \text{(Option A)}
]
Explanation:
The problem involves understanding molality, which focuses on the solvent’s weight rather than the solution’s total volume. The given density of THF allows us to convert its volume into mass, a critical step since molality requires the solvent’s mass in kilograms. After determining the moles of dichlorobenzene using its molar mass, the calculated molality is 0.903 mol/kg, reflecting the solute concentration in the solution.