Question

$58.4\, gm$ of $NaCl$ and $180\,gm$ of glucose were separately dissolved in $1000 \,mL$ of water. Identify the correct statement regarding the elevation of boiling point ($b.pt$) of the resulting solutions.

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

Answer 1

Answer: (A)

NaCl solution will show higher elevation of b.pt

Answer 2

Elevation in boiling point, $\Delta T_{b}=i \times K_{b} \times m$
Molality of $NaCl$ solution $=\frac{n}{W} \times 1000$
$=\frac{\frac{58.5}{58.5}}{W_{ H _{2} O }} \times 1000$
$=\frac{1000}{W_{ H _{2}} O }$
Molality of $C _{6} H _{12} O _{6}$ solution $=\frac{\frac{180}{180} \times 1000}{W_{ H _{2}} O }$
$=\frac{1000}{W_{ H _{2}} O }$
Both the solutions have same molarity but the values for $NaCl$ and glucose are 2 and 1 respectively.
$\therefore \Delta T_{b( NaCl )}$
$=2 \times \Delta T_{b\left( C _{6} H _{12} O _{6}\right)}$