Question

what is the new boiling point of a solution prepared by adding 296.0 g of sucrose to 383 g of water? the boiling point constant for water is 0.52 °c/m. molar mass of sucrose is 342.30 g/mol hint: convert mass of solute to moles and grams of solvent to kilograms. 130.3 °c 401.9 °c 100.4 °c 101.2 °c

Explanation:

Step1: Calculate moles of sucrose

Moles of sucrose = mass / molar mass = \( \frac{296.0\ g}{342.30\ g/mol} \approx 0.8647\ mol \)

Step2: Convert solvent mass to kg

Mass of water (solvent) = \( 383\ g = 0.383\ kg \)

Step3: Calculate molality (m)

Molality \( m = \frac{moles\ of\ solute}{kg\ of\ solvent} = \frac{0.8647\ mol}{0.383\ kg} \approx 2.258\ m \)

Step4: Calculate boiling point elevation (\(\Delta T_b\))

\(\Delta T_b = K_b \times m\), where \( K_b = 0.52\ ^\circ C/m \)
\(\Delta T_b = 0.52\ ^\circ C/m \times 2.258\ m \approx 1.174\ ^\circ C \)

Step5: Find new boiling point

Boiling point of pure water is \( 100^\circ C \), so new boiling point = \( 100^\circ C + 1.174^\circ C \approx 101.2^\circ C \)

Answer:

101.2 \(^\circ\)C (Option: 101.2 \(^\circ\)C)