Question #345558

**What is the molality of para-dichlorobenzene in a solution prepared by dissolving 2.65g C6H4Cl2 in 50.0 mL of benzene (d=0.897g/mL)?**

Expert's answer

**Solution:**

The solute is para-dichlorobenzene (C_{6}H_{4}Cl_{2})

The solvent is benzene (C_{6}H_{6})

Density = Mass / Volume

Therefore,

Mass of C_{6}H_{6} = Density × Volume = (0.897 g mL^{−1}) × (50.0 mL) = 44.85 g

Kilograms of C_{6}H_{6} = (44.85 g) × (1 kg / 1000 g) = 0.04485 kg

Moles = Mass / Molar mass

The molar mass of para-dichlorobenzene (C_{6}H_{4}Cl_{2}) is 147 g mol^{−1}

Therefore,

Moles of C_{6}H_{4}Cl_{2} = (2.65 g) / (147 g mol^{−1}) = 0.01803 mol

Molality = Moles of solute / Kilograms of solvent

Therefore,

Molality of C_{6}H_{4}Cl_{2 }solution = Moles of C_{6}H_{4}Cl_{2} / Kilograms of C_{6}H_{6}

Molality of C_{6}H_{4}Cl_{2 }solution = (0.01803 mol) / (0.04485 kg) = 0.402 mol kg^{−1} = 0.402 m

**Molality of C**_{6}**H**_{4}**Cl**_{2 }**solution = 0.402 m**

**Answer: The molality of para-dichlorobenzene (C**_{6}**H**_{4}**Cl**_{2}**) solution is 0.402 m**

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