TS EAMCET 5 May 2018 Shift 1 Solved Paper

The reaction occurring is as shown below.
CaCO3

2KOH

––––––––🢖

Δ

CO2
This implies that 2 moles of KOH are required to neutralize 1mole of CO2.
The molar mass of KOH given is 56 g.
The mass of KOH will be 112(56×2) since 2 moles are reacting.
112g of NaOH will neutralize 100g of CaCO3
So 28g of KOH will neutalize =

100×28

112

=25g of CaCO3
Now, 60 g of impure CaCO3, has 25g of pure CaCO3.
So, the percentage purity present in 100 g of impure CaCO3 iscalculated as follows:
Percentage purity =

25×100

60

=41.6g