Calculate the molarity of the following solution:
(a) 4g of caustic soda is dissolved in 200mL of the solution.
(b) 5.3 g of anhydrous sodium carbonate is dissolved in100 mL of solution.
(c ) 0.365 g of pure HCl gas is dissolved in 50 mL of solution.

Explanation

(a) Molecular mass of NaOH = 40
No. of gram moles in 4g of NaOH `= (4)/(40)=0.1`
Volume of solution in litre `= (200)/(1000)=0.2`
Molarity `= ("No. of moles of solute")/("Volume of solution in litre")=(0.1)/(0.2)=0.5M`
(b) Molecular mass of anhydrous `Na_(2)CO_(3)=106`
No. of gram moles in 5.3 g of anhydrous `Na_(2)CO_(3)`
`=(5.3)/(106)=0.05`
Volume of solution in litre `= (100)/(1000)=0.1`
Molarity `= ("No. of gram moles of solute")/("Volume of solution in litre")`
`=(0.05)/(0.1)=0.5M`
(c) Molarity mass of HCl = 36.5
No. of gram moles in 0.365g of pure HCl `= (0.365)/(36.5)`
Volume of solution in litre `= (50)/(1000)`
Molarity `= ("No. of gram moles of solute")/("Volume of solution in litre")`
`=(0.365)/(36.5)xx(1000)/(50)=0.2M`.