Question

Which of the following solutions will have the highest boiling point ?

• A. $0.1 \,M \,FeCl_3$
• B. $0.1 \,M\, BaCl_2$
• C. $0.1 \,M \,NaCl$
• D. $0.1 M\, urea \,(NH_2CONH_2)$

Answer 1

Answer: (A)

$0.1 \,M \,FeCl_3$

Answer 2

Except for urea (a non electrolyte) the ionisation of $FeCl_3, BaCl_2, NaCl$ can be represented as follows ${FeCl_3-> Fe^{3+} + 3Cl^-}$ ${ BaCl_2 -> Ba2+ + 2Cl^-}$ ${NaCl -> Na+ + Cl^-}$ As number of particles (ions) produced is maximum in $FeCl_3$ the elevation in boiling point (a colligative property) will be maximum in case of $FeCl_3$.