• # question_answer ${{H}_{2}}O.$ Mole fraction of $NaOH$ in solution and molality $(in\,mol\,k{{g}^{-1}})$ of the solution respectively are: A) $0.2,\text{ }22.\text{ }20$ B) $0.2,\text{ }11\text{ }.11$ C) $0.167,\text{ }11.\text{ }11$ D) $0.167,\text{ }22.20$

 Moles of   $NaOH=\frac{8}{40}=0.2$ Moles of ${{H}_{2}}O=\frac{18}{18}=1$ Mole fraction of $NaOH=\frac{0.2}{1.2}=0.167$ Molality $=\frac{8}{40}\times \frac{1000}{18}=11.11.$