Elevation in boiling point, $\mathrm{\Delta }{T}_b=i\times K_b\times m$
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 $f{\mathrm{C}}_6{\mathrm{H}}_1\text{⧸}{\mathrm{O}}_6$ 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. $\therefore \mathrm{\Delta }{T}_{b(\mathrm{NaG})}=2\times \mathrm{\Delta }{T}_{bC{C}_6{H}_{1}2}{\mathrm{O}}_6)$