How much of sucrose is to be added to 500 g of water such that it boils at 100°C if the molar elevation constant for water is 0.52 K kg mol-1 and the boiling point of water at 750 mm Hg is 99.63°C?

Elevation in boiling point $\Delta T_{b}=(100+273)-(99.63+273)$

$=0.37 \mathrm{~K}$

Mass of water, $w_{1}=500 \mathrm{~g}$

Molar mass of sucrose $\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right), M_{2}=11 \times 12+22 \times 1+11 \times 16$

$=342 \mathrm{~g} \mathrm{~mol}^{-1}$

Molar elevation constant, $\mathrm{K}_{\mathrm{b}}=0.52 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$

Now we have:

$\Delta {{T}_{b}}=\frac{{{K}_{b}}\times 1000\times {{w}_{2}}}{{{M}_{2}}\times {{w}_{1}}}$

$\Rightarrow {{w}_{2}}=\frac{\Delta {{T}_{b}}\times {{M}_{2}}\times {{w}_{1}}}{{{K}_{b}}\times 1000}$

$=\frac{0.37\times 342\times 500}{0.52\times 1000}$

$=121.67~\text{g (approximately) }$ As a result, the amount of sucrose that is to be added is $121.67 \mathrm{~g}$