Question

$58.5g$ of $NaCl$ and $180g$ 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 boiling point
• B. Glucose solution will show higher elevation of boiling point
• C. Both the solutions will show equal elevation of boiling point
• D. The boiling point elevation will be shown by neither of the solutions

Answer 1

The correct option is A $NaCl$ solution will show higher elevation of boiling point

Elevation in boiling point, $△T_b=i\times K_b\times m$

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 "$i$" for $NaCl$ and glucose are $2$ and $1$ respectively.

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