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Multiple Choice
A 3.60 m aqueous sucrose solution is prepared. Given that the boiling point elevation constant (K_b) for water is 0.512 °C·kg/mol and pure water boils at 100.00 °C, what is the boiling point of the solution? Assume sucrose does not ionize in water.
A
100.51 °C
B
101.84 °C
C
100.00 °C
D
102.56 °C
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Verified step by step guidance
1
Identify the given information: molality (m) = 3.60 mol/kg, boiling point elevation constant (K_b) = 0.512 °C·kg/mol, and the normal boiling point of pure water = 100.00 °C.
Recall the formula for boiling point elevation: \(\Delta T_b = i \times K_b \times m\), where \(\Delta T_b\) is the boiling point elevation, \(i\) is the van't Hoff factor, \(K_b\) is the ebullioscopic constant, and \(m\) is the molality.
Since sucrose does not ionize in water, the van't Hoff factor \(i\) equals 1.
Calculate the boiling point elevation \(\Delta T_b\) by substituting the values: \(\Delta T_b = 1 \times 0.512 \times 3.60\).
Add the boiling point elevation to the normal boiling point of water to find the boiling point of the solution: \(T_{boiling} = 100.00 + \Delta T_b\).