Question

10 g impure CaCO3 (lime stone) is heated to give CO2 which is allowed to pass in water to form H2CO3. This solution is completely neutralized by

500 ml of 0.2 M KOH solution. The % purity of lime

stone is​

Answer 1

Answer:CaCO  

3

​  

 

⇌

Δ

​  

CaO+CO  

2

​  

 

Let pure sample of CaCO  

3

​  

=x grams

Mass of CaO produced after decomposition =22.4 g

Molar mass of CaCO  

3

​  

=100 g/mol

and, Molar mass of CaO=56 g

If 100% is pure, then

100 g CaCO  

3

​  

⟶56 g of CaO

Also,y g of CaCO  

3

​  

⟶22.4 g of CaO

Dividing these two,

y

100

​  

=  

22.4

56

​  

 

y=40 grams

∴ Percentage of purity =  

Totalmassofimpure

Massofpuresample

​  

×100=  

50

40

​  

×100=80%

Explanation: