Calculate the molarity of the following solution:
(a) 4 g of caustic soda is dissolved in 200 mL of the solution.
(b) 5.3 g of anhydrous sodium carbonate is dissolved in 100 mL of solution.
(c) 0.365 g of pure HCl gas is dissolved in 50 mL of solution.

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Solution

(a) Molecular mass of NaOH = 40
No. of gram moles in 4 g of NaOH = $\frac{4}{40} = 0.1$
Volume of solution in litre = $\frac{200}{1000} = 0.2$
$M o l a r i t y = \frac{N o . o f m o l e s o f s o l u t e}{V o l u m e o f s o l u t i o n i n l i t r e} = \frac{0.1}{0.2} = 0.5 M$
(b) Molecular mass of anhydrous $N a_{2} C O_{3} = 106$
No. of gram moles in 5.3 g of anhydrous $N a_{2} C O_{3}$
$= \frac{5.3}{106} = 0.05$
Volume of solution in litre $= \frac{100}{1000} = 0.1$
Molarity $= \frac{N o . o f g r a m m o l e s o f s o l u t e}{V o l u m e o f s o l u t i o n i n l i t r e} = \frac{0.05}{0.1} = 0.5 M$
(c) Molecular mass of HCl = 36.5
No. of gram moles in 0.365 g of pure HCl $= \frac{0.365}{36.5}$
Volume of solution in litre $= \frac{50}{1000}$
$M o l a r i t y = \frac{N o . o f g r a m m o l e s o f s o l u t e}{V o l u m e o f s o l u t i o n i n l i t r e}$
$= \frac{0.365}{36.5} \times \frac{1000}{50} = 0.2 M$

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