Elevation in boiling point of a molar (1 M) glucose solution $(\mathrm{d}=1.2\mathrm{g}{\mathrm{mL}}^{-1})$ is

Molarity = 1 M = 1 mol ${\mathrm{L}}^{-1}$

1000 mL of solution has glucose = 180 g

$(1000\times 1.2)$g of solution has glucose = 180 g

$\therefore$ Solute $\left({\mathrm{w}}_1\right)=180\mathrm{g}$

Solution = 1200 g

Solvent, ${\mathrm{H}}_2\mathrm{O}\left({\mathrm{w}}_2\right)=1200-180=1020 \mathrm{g}$

$\therefore {\mathrm{\Delta T}}_{\mathrm{b}}=\frac{1000{\mathrm{K}}_{\mathrm{b}}{\mathrm{w}}_1}{{\mathrm{m}}_1{\mathrm{w}}_2}$

         $$\begin{gathered} =\frac{1000\times {\mathrm{K}}_{\mathrm{b}}\times 180}{180\times 1020} \\ =0.98{\mathrm{K}}_{\mathrm{b}} \end{gathered}$$