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Step-by-Step Solution
Step 1: Identify the Known Data
• Pure solvent (Benzene) freezing point, T_f^o = 5.5^\circ C .
• Freezing point depression constant of benzene, K_f = 5.12^\circ C/m .
• Mass of solute (Butane, C_4H_{10} ), m_\text{solute} = 10\,\text{g} .
• Molar mass of solute (Butane, C_4H_{10} ) = 12\times4 + 10 = 58\,\text{g/mol} .
• Mass of solvent (Benzene, C_6H_6 ), m_\text{solvent} = 200\,\text{g} = 0.2\,\text{kg} .
• Since C_4H_{10} does not dissociate, the van ’t Hoff factor i = 1 .
Step 2: Calculate the Number of Moles of Solute
Number of moles of solute, n_\text{solute} =
\displaystyle \frac{\text{mass of solute}}{\text{molar mass of solute}}
= \frac{10}{58}
= 0.1724\,\text{mol (approximately)}.
Step 3: Calculate the Molality of the Solution
Molality ( m ) =
\displaystyle \frac{n_\text{solute}}{\text{mass of solvent in kg}}
= \frac{0.1724}{0.2}
= 0.862\,\text{mol/kg (approximately)}.
Step 4: Apply the Freezing Point Depression Formula
The freezing point depression is given by
\Delta T_f = i \times K_f \times m
.
Here,
\Delta T_f = 1 \times 5.12^\circ C/m \times 0.862 \approx 4.415^\circ C.
Step 5: Determine the New Freezing Point
T_f = T_f^o - \Delta T_f
= 5.5^\circ C - 4.415^\circ C
= 1.085^\circ C.
Thus, the solution freezes at approximately 1.09^\circ C (often rounded to 1^\circ C ).
