100 ml of N/5 NaOH will neutralize:

Explanation

The correct answer is 1.2368g of H3BO3. The reaction between NaOH and H3BO3 is a neutralization reaction, where the base (NaOH) reacts with the acid (H3BO3) to form water and a salt. Using the normality of NaOH (N/5 or 0.2N) and the volume (100 ml or 0.1 L), we calculate the equivalent moles of NaOH, which is 0.02 equivalents. H3BO3 has a molar mass of 61.83 g/mol, and since it provides one equivalent per mole in the reaction, the equivalent weight is the same as its molar mass. Thus, 0.02 equivalents of H3BO3 corresponds to 0.02 x 61.83 = 1.2368 g, which can be neutralized by 100 ml of N/5 NaOH. The other options do not align with this stoichiometric requirement.