The elevation of boiling point of water produced by dissolving 1.17 g sodium chloride in 100 g water $\left(K_{\mathrm{b}}=0.512\right.\left.{\mathrm{Kkgmol}}^{-1}\right)$ is

Molality of sodium chloride $=\frac{(1.17/58.5)\mathrm{mol}}{0.1\mathrm{kg}}=0.2\mathrm{mol}{\mathrm{kg}}^{-1}$

Molality of ions in solution $=2\times 0.2\mathrm{mol}{\mathrm{kg}}^{-1}\mathrm{\Delta }{T}_{\mathrm{b}}=K_{\mathrm{b}}m=\left(0.512\mathrm{Kkg}{\mathrm{mol}}^{-1}\right)\left(2\times 0.2\mathrm{mol}{\mathrm{kg}}^{-1}\right)=0.205\mathrm{K}$