4.0 g of NaOH contained in one deciliter of a solution. Calculate the following in this solution:
(i) Mole fraction of NaOH
(ii) Molality of NaOH
(iii) Molality of NaOH
(iii) Molarity of NaOH.
(At. wt. of Na = 23, O = 16; Density of NaOH solution is 1.038 $g / c m^{3}$ ).

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Solution

Mass = $d e n s i t y \times v o l u m e$
Mass of 0.1 l NaOH solution $= 1.038 g / c m^{3} \times 100 c m^{3} = 103.8 g$
1) 4 g of NaOH is present in 103.8 g of solution.
Therefore solvent is 99.8 g.
Number of moles of NaOH =
$\frac{4}{40} = 0.1 m o l$
Number of moles of water =
$\frac{99.8}{18} = 5.44 m o l$
Total number of moles = 0.1 + 5.544 = 5.644 moles
Mole fraction of NaOH $= \frac{N u m b e r o f m o l e s o f N a O H}{T o t a l n u m b e r o f m o l e s} = \frac{0.1}{5.644} = 0.0177$
2) Molality $= \frac{N u m b e r o f m o l e s o f N a O H \times 1000}{w e i g h t o f s o l v e n t i n g} = \frac{0.1 \times 1000}{99.8} = 1.002 m o l / k g$
3) Molarity $= \frac{N u m b e r o f m o l e s o f N a O H}{v o l u m e o f s o l u t i o n i n 1} = \frac{0.1}{0.1} = 1 m o l / l$

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