Aqueous solution of $N a O H$ is marked $10 % (w / w) .$ The density of the solution is $1.070 {g c m}^{- 3}$ Calculate:
(i) molarity
(ii) molality and
(iii) mole fraction of $N a O H$ $&$ water.
[ $N a = 23, H = 10, O = 16$ ]

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Solution

Here if an 100 g of an diluted $N a O H$ solution is marked as $10 % (w / w)$ of $N a O H$ then $W_{2} (mass of solute) = 10 g$ and mass of water $(W_{1}) = 90 g$ .

$d e n s i t y = \frac{m a s s}{V_{s o l u t i o n}}$

$V_{s o l u t i o n} = \frac{100}{1.070} = 93.46 c m^{- 3} = 0.09346 L$

number of moles of $N a O H (n_{s o l u t e}) = \frac{10 g}{49 g (m o l)^{- 1}} = 0.2 m o l$
$M o l a r i t y = \frac{n_{s o l u t e}}{V_{s o l u t i o n}}$

$M o l a r i t y = \frac{0.2 m o l}{0.09346 L} = 2.14 M$

$m o l a l i t y = \frac{n_{s o l u t e}}{W_{1} (i n k g)}$

$m o l a l i t y = \frac{0.2 m o l}{0.09 k g} = 2.22 m$

Let mole fraction of water and $N a O H$ be $x_{1}$ and $x_{2}$
and $n_{1}$ and $n_{2}$ be the moles of water and $N a O H$

$n_{1} = 0.2 m o l$ and $n_{2} = \frac{90 g}{18 g (m o l)^{- 1}} = 5 m o l$

$x_{1} = \frac{n_{1}}{n_{1} + n_{2}} = 0.96$

$x_{2} = \frac{n_{2}}{n_{1} + n_{2}} = 0.04$

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