how many cl minus ions are present in 100ml of 0.2M cacl2 aqeous solution ofNaOH , the density of which is 1.1g/ml.also find the mole fraction of NaOH in this solution

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Solution

$M i l l i m o l o f C a C l_{2} = 100 \times 0.2 = 20 m m o l C a C l_{2} \to C a^{2 +} + 2 C l^{-} 20 m m o l o f C a C l_{2} = 2 \times 20 m m o l o f C l^{-} = 40 m m o l D e n s i t y o f N a O H = 1.1 g / m l W e i g h t o f N a O H = 1.1 g V o l u m e o f W a t e r = 1 m l = 1 g N u m b e r o f m o l e s = \frac{G i v e n m a s s}{M o l a r m a s s} N u m b e r o f m o l e s o f N a O H (n_{2}) = \frac{1.1}{40} = 0.0275 N u m b e r o f m o l e s o f H_{2} O (n_{1}) = \frac{1}{18} = 0.0555 M o l e f r a c t i o n o f N a O H = \frac{n_{2}}{n_{1} + n_{2}} = \frac{0.0275}{0.0555 + 0.0275} = 0.3313$

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