2.0g of a monobasic acid was made up to 250 cm3 with distilled water. 25.00 cm3 of this solution required 20.00cm3 of 0.1 M NaOH solution for complete neutr...

Answer Details

This is a question on acid-base titration. From the given data, we know that 25.00 cm³ of the monobasic acid requires 20.00 cm³ of 0.1 M NaOH solution for complete neutralization. Therefore, the number of moles of NaOH required to neutralize the acid is: 0.1 mol/L × 0.020 L = 0.002 mol Since the acid is monobasic, the number of moles of the acid is also 0.002 mol. We can use this information to calculate the molar mass of the acid. The molar mass is equal to the mass of the acid divided by the number of moles of the acid: Molar mass = Mass of acid / Number of moles of acid We are given that 2.0 g of the acid was made up to 250 cm³. Therefore, the concentration of the acid is: Concentration = Mass of acid / Volume of solution = 2.0 g / 250 cm³ = 0.008 mol/L Now, we can calculate the number of moles of the acid present in 25.00 cm³ of the solution: Number of moles of acid = Concentration × Volume = 0.008 mol/L × 0.0250 L = 0.0002 mol Finally, we can calculate the molar mass of the acid: Molar mass = Mass of acid / Number of moles of acid = 2.0 g / 0.0002 mol = 100 g/mol Therefore, the molar mass of the acid is 100 g/mol. Answer option (C) is correct.