Question

$58.4gm$ of $NaCl$ and $180gm$ of glucose were separately dissolved in $1000mL$ 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)

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.
$\therefore \Delta T_{b(NaCl)}$
$=2\times \Delta T_{b\left(C_6{H}_{12}{O}_6\right)}$