Question

Which of the following solutions will have a maximum boiling point?

• A. 1% glucose in water
• B. 1% \(CaCl_2\) in water
• C. 1% sucrose in water
• D. 1% NaCl in water

Hint:

Boiling point rises with \(i\times\) molality. For equal mass %, compare \(i/M\); the smallest-mass strong electrolyte wins.

Answer 1

The Correct Option is D

Step 1: Concept. Elevation of boiling point is a colligative property: \[\Delta T_b = i\,K_b\,m\] where \(i\) is the van't Hoff factor (number of particles a solute gives on dissolving) and \(m\) is molality. So the solution that produces the largest number of dissolved particles boils highest.
Step 2: Same mass basis. All four are 1% by mass, so each has the same mass of solute in the same water. Moles \(=\) mass/M, hence molality \(\propto 1/M\). The effective particle count is therefore \(\propto i/M\).
Step 3: Compare \(i/M\).
Glucose: \(i=1,\ M=180 \Rightarrow 0.0056\)
\(CaCl_2\): \(i=3,\ M=111 \Rightarrow 0.027\)
Sucrose: \(i=1,\ M=342 \Rightarrow 0.0029\)
NaCl: \(i=2,\ M=58.5 \Rightarrow 0.0342\)
Step 4: Decide. NaCl gives the largest \(i/M\), so the most particles and the highest \(\Delta T_b\).
Why others are wrong: Glucose and sucrose are non-electrolytes (\(i=1\)), so few particles. \(CaCl_2\) gives 3 ions but its high molar mass makes its particle count fall just below NaCl.
\[\boxed{\text{1% NaCl in water (option iv)}}\]