The boiling point of water at 735 Torr is 99.07 °C. The mass of $\mathrm{NaCl}$ added in 100 g water $\left(K_{\mathrm{b}}=0.51\mathrm{Kkg}{\mathrm{mol}}^{-1}\right)$ to make its boiling point 100 °C is

$m=\frac{\mathrm{\Delta }{T}_{\mathrm{b}}}{K_{\mathrm{b}}}=\frac{(100-99.07)}{0.51}\mathrm{mol}{\mathrm{kg}}^{-1}$

Assuming$\mathrm{NaCl}$ to be completely dissociated, we will have

Molality of $\mathrm{NaCl}=\frac{1}{2}\left(\frac{100-99.07}{0.51}\mathrm{mol}{\mathrm{kg}}^{-1}\right)$

Mass of $\mathrm{NaCl}$in 100 g water $=\left(\frac{1}{10}\right)\frac{1}{2}\left(\frac{0.93}{0.51}\right)\left(58.5\right)\mathrm{g}=5.33\mathrm{g}$