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$58.4 \mathrm{g}$ of $\mathrm{NaCl}$ and $180 \mathrm{g}$ of glucose were separately dissolved in $1000 \mathrm{mL}$ of water. Identify the correct statement regarding the elevation of boiling point (b.p) of the resulting solutions.
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The correct answer is:
NaCl solution will show higher elevation of boiling point
Elevation in boiling point, $\Delta T_{b}=i \times K_{b} \times m$
Molality of NaCl solution $=\frac{n}{W} \times 1000$
$=\frac{58.5}{W_{\mathrm{H}_{2} \mathrm{O}}} \times 1000=\frac{1000}{\mathrm{W}_{\mathrm{M}_{2}} \mathrm{O}}$
Molality of $f \mathrm{C}_{6} \mathrm{H}_{1} \not \mathrm{O}_{6}$ solution = $\frac{\frac{180}{180} \times 1000}{w_{\mathrm{H}_{2}} \mathrm{O}}$
$\frac{1000}{W_{M_{2}} O}$. Both the solutions have same molarity but the values for $\mathrm{NaCl}$ and glucose are ${2}$ and 1respectively. $\left.\therefore \quad \Delta T_{b(\mathrm{NaG})}=2 \times \Delta T_{b C C_{6} H_{1} 2} \mathrm{O}_{6}\right)$
Molality of NaCl solution $=\frac{n}{W} \times 1000$
$=\frac{58.5}{W_{\mathrm{H}_{2} \mathrm{O}}} \times 1000=\frac{1000}{\mathrm{W}_{\mathrm{M}_{2}} \mathrm{O}}$
Molality of $f \mathrm{C}_{6} \mathrm{H}_{1} \not \mathrm{O}_{6}$ solution = $\frac{\frac{180}{180} \times 1000}{w_{\mathrm{H}_{2}} \mathrm{O}}$
$\frac{1000}{W_{M_{2}} O}$. Both the solutions have same molarity but the values for $\mathrm{NaCl}$ and glucose are ${2}$ and 1respectively. $\left.\therefore \quad \Delta T_{b(\mathrm{NaG})}=2 \times \Delta T_{b C C_{6} H_{1} 2} \mathrm{O}_{6}\right)$
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