If $4 \ g$ of $NaOH$ dissolves in $36 \ g$ of $H_2O$,calculate the mole fraction of each component in the solution. Also,determine the molarity of the solution (specific gravity of the solution is $1 \ g \ mL^{-1}$).

Mass of $NaOH = 4 \ g$
Number of moles of $NaOH = \frac{4 \ g}{40 \ g \ mol^{-1}} = 0.1 \ mol$
Mass of $H_2O = 36 \ g$
Number of moles of $H_2O = \frac{36 \ g}{18 \ g \ mol^{-1}} = 2 \ mol$
Mole fraction of water $(x_{H_2O})$ $= \frac{2}{2 + 0.1} = \frac{2}{2.1} \approx 0.952$
Mole fraction of $NaOH$ $(x_{NaOH})$ $= \frac{0.1}{2 + 0.1} = \frac{0.1}{2.1} \approx 0.048$
Mass of solution $= 36 \ g + 4 \ g = 40 \ g$
Volume of solution $= \frac{\text{Mass}}{\text{Density}} = \frac{40 \ g}{1 \ g \ mL^{-1}} = 40 \ mL = 0.04 \ L$
Molarity of solution $= \frac{\text{moles of solute}}{\text{Volume of solution in } L} = \frac{0.1 \ mol}{0.04 \ L} = 2.5 \ M$