Question

A 0.1 molal aqueous solution of glucose boils at \( 100.16^\circ \text{C} \). The boiling point of a 0.5 molal aqueous solution of glucose will be:

• A. \( 100.56^\circ \text{C} \)
• B. \( 100.40^\circ \text{C} \)
• C. \( 100.80^\circ \text{C} \)
• D. \( 101.00^\circ \text{C} \)

Hint:

Colligative properties depend only on the number of solute particles in the solution, not their identity.

Answer 1

The Correct Option is C

The boiling point elevation (\( \Delta T_b \)) is given by: \[ \Delta T_b = i K_b m, \] where: \begin{itemize}
\( i \) is the van’t Hoff factor (for glucose, \( i = 1 \)),
\( K_b \) is the boiling point elevation constant,
\( m \) is the molality of the solution. \end{itemize} From the data: \[ \Delta T_b (\text{for 0.1 m}) = 100.16 - 100 = 0.16^\circ \text{C}. \] The proportionality of \( \Delta T_b \) with molality gives: \[ \Delta T_b (\text{for 0.5 m}) = 5 \times 0.16 = 0.80^\circ \text{C}. \] Thus, the new boiling point is: \[ 100 + 0.80 = 100.80^\circ \text{C}. \] ---
Final Answer: \[ \boxed{100.80^\circ \text{C}} \]