Calculation of % (w/w) CaCO3 in the sample

Given:
- 1g of chalk is completely neutralised by 100 ml 0.04N HCl solution.

The molar concentration of HCl solution can be calculated as follows:
- N = Normality of the solution = 0.04
- Molar mass of HCl = 36.5 g/mol
- N = (No. of H+ ions) x Molarity = (No. of H+ ions) x (Concentration in g/L) / Molar mass
- No. of H+ ions = N x Molar mass / Concentration in g/L = 0.04 x 36.5 / 1000 = 0.00146 mol/L

The amount of HCl required to neutralise 1g of chalk can be calculated as follows:
- 1 mole of CaCO3 reacts with 2 moles of HCl
- Therefore, the amount of HCl required to neutralise 1g of CaCO3 = 2 x 0.00146 = 0.00292 moles
- The amount of HCl present in 100 ml of 0.04N HCl solution = 0.04 x 36.5 / 1000 x 100 = 0.146 moles
- Therefore, the amount of CaCO3 present in 1g of chalk = 0.00292 moles
- % (w/w) CaCO3 in the sample = (Mass of CaCO3 / Total mass of sample) x 100
- Total mass of sample = 1g
- Mass of CaCO3 = Molar mass of CaCO3 x Amount of CaCO3 present = 100 x 0.00292 = 0.292g
- % (w/w) CaCO3 in the sample = (0.292 / 1) x 100 = 29.2%

Therefore, the correct answer is not provided in the options given. The closest option is 20, but the correct answer is actually 29.2%.

20%