A 0.4 g sample of monobasic acid requires 16cm3 of 0.1 mol per dm3 of NaOH for complete neutralization . what is the relative mass of the acid?? Plz ans me urgent
HA + NaOH ==> NaA + H2O
mols NaOH = M x L = ?
mols HA = mols NaOH (look at the 1:1 coefficients)
mols HA = grams/molar mass
You know mols HA and grams, solve for molar mass.
To find the relative mass of the acid, we need to use the concept of molar ratios. Here's how to solve the problem:
Step 1: Calculate the number of moles of NaOH used.
We are given the volume of NaOH solution used, which is 16 cm³. However, we need to convert this to dm³. Since 1 dm³ = 1000 cm³, the volume of NaOH solution in dm³ is 16/1000 = 0.016 dm³.
The concentration of the NaOH solution is given as 0.1 mol/dm³. Therefore, the number of moles of NaOH used can be calculated using the formula:
Moles of NaOH = Concentration × Volume
Moles of NaOH = 0.1 mol/dm³ × 0.016 dm³
Moles of NaOH = 0.0016 mol
Step 2: Determine the molar ratio between NaOH and the monobasic acid.
The balanced chemical equation for the neutralization reaction between an acid (HA) and a base (NaOH) is:
HA + NaOH → NaA + H₂O
From the equation, we can see that one mole of acid reacts with one mole of NaOH. Therefore, the molar ratio between NaOH and the monobasic acid is 1:1.
Step 3: Calculate the number of moles of the monobasic acid.
Since the molar ratio is 1:1, the number of moles of the monobasic acid is also 0.0016 mol.
Step 4: Calculate the relative mass of the acid.
The relative mass is calculated by dividing the mass of the substance by the number of moles. In this case, the mass of the acid is given as 0.4 g. Therefore:
Relative mass of the acid = Mass of the acid / Moles of the acid
Relative mass of the acid = 0.4 g / 0.0016 mol
Relative mass of the acid = 250 g/mol
Thus, the relative mass of the monobasic acid is 250 g/mol.