A 3.664 g sample of monoprotic acid was dissolved in water and required 20.27 mL of a 0.1578 M NaOH solution for neutralization . Calculate the molar mass of the acid
moles acid=moles base
3.664/molmass=1.578*.02027
solve for molemass
To calculate the molar mass of the monoprotic acid, we need to use the concept of stoichiometry.
First, let's calculate the number of moles of NaOH used in the neutralization reaction. We can use the formula:
moles of NaOH = concentration of NaOH (in M) × volume of NaOH (in L)
Given that the concentration of NaOH is 0.1578 M and the volume of NaOH is 20.27 mL (which is equivalent to 0.02027 L), we can substitute these values into the equation:
moles of NaOH = 0.1578 M × 0.02027 L
moles of NaOH ≈ 0.003193 mol
Since NaOH and the monoprotic acid react in a 1:1 ratio, the number of moles of the acid is also 0.003193 mol.
Next, we can calculate the molar mass of the acid. Molar mass is defined as the mass of one mole of a substance. We can use the formula:
molar mass = mass of the sample (in g) / moles of the substance
Given that the mass of the sample is 3.664 g and the number of moles of the acid is 0.003193 mol, we can substitute these values into the equation:
molar mass = 3.664 g / 0.003193 mol
molar mass ≈ 1146.3 g/mol
Therefore, the molar mass of the monoprotic acid is approximately 1146.3 g/mol.