What mass of 95% pure CaCO₃ will be required to neutralise 50 mL of 0.5 MHCl solution according to the following reaction?

CaCO_3(s) +2HCl_(aq) → CaCl_2(aq) + CO_2(g) + 2H₂O_(l)

[Calculate upto second place of decimal point]

Explanation:

Mass percent - It is a concentration term used to determine the strength of the solution. It is defined as the total percentage of the mass of solute present in the given solution.In the given reaction - 

CaCO3(s) +2HCl(aq) → CaCl2(aq) + CO2(g) + 2H2O(l)

The number of moles of pure CaCO3 is half of the number of moles of HCl;

So, the number of moles of pure CaCO3 = 

Moles of HCl = molarity of HCl × volume of HCl in liters (∵ molarity = )

= 0.5 × 

= 0.5 × 0.05 = 0.025

So, the number of moles of pure CaCO₃ is = 

=0.0125

weight of pure CaCO3 is = number of moles of CaCO3 pure × molecular wt. of CaCO3 

= 0.0125 × 100     (∵ mol wt. of CaCO3 is 100)

=1.25g

Percentage of purity of CaCO3 in sample = 

given purity = 95%

So, 95% = 

total wt. of sample(impure) =  = 1.3157 ∼ 1.32 g 

So, a 1.32 g sample of CaCO₃ is required to neutralize the given HCl.

Hence, the correct answer is option 3.