Question

$58.5gm$ of $NaCl$ and $180gm$ of glucose were separately dissolved in $1000ml$ of water. Identify the correct statement regarding the elevation of boiling point (b.p.) of the resulting solutions.

• A. $NaCl$ solution will show higher elevation of b.p.
• B. Glucose solution will show higher elevation of b.p.
• C. Both the solutions will show equal elevation of b.p.
• D. The b.p. elevation will be shown by neither of the solutions

Answer 1

The correct option is A $NaCl$ solution will show higher elevation of b.p.

The elevation of boiling point (b.p.)
$\Delta T_b=i{K}_{b}m$
$\Delta T_b\propto i\times m$
The soultion having higher $i\times m$ will have higher $\Delta T_b$
For NaCl solution $i=2$ as one molecule of NaCl dissociates to produce two ions.
The molality of NaCl $m=\frac{\text{58.5 g}}{\text{58.5 g/mol}\times \text{1000 mL}\times \text{1 g/mL}}\times \text{1000 g/kg}=\text{1 m}$
For NaCl solution $i\times m=2\times 1=2$
For glucose solution $i=1$ as molecule of glucose does not dissociates.
The molality of glucose $m=\frac{\text{180 g}}{\text{180 g/mol}\times \text{1000 mL}\times \text{1 g/mL}}\times \text{1000 g/kg}=\text{1 m}$
For glucose solution $i\times m=1\times 1=1$
Thus, NaCl solution will have higher elevation of boiling point ($\Delta T_b$) than glucose solution.