.4g sample of monobasic acid requires 16cm3 of .1 mol dm 3 of NAOH for complete neutralization what is relative mass of acid
Monobasic acid has only 1 replaceable hydrogen atom:
Thus, let it be rep as HX:
Equation of rxn:
HX + NaOH ------> NaX + H2O
Mole ratio: na/nb = 1/1, then;
CaVa = CbVb
Vb = 16cm3 = 0.016dm3
Cb = 0.1mol/dm3
Mass of acid = 0.4g
Where CbVb = Nb,
0.016 x 0.1 = 0.0016moles
Since na = nb,
0.0016moles = na
Nos of moles = used mass/molar mass
0.0016 = 0.4/molar mass
Molar mass = 0.4/0.0016 = 250g/mol
= 250g/mol
250
To find the relative mass of the monobasic acid, we need to use the information given in the question.
Step 1: Calculate the number of moles of NaOH used.
We are given that 16 cm^3 (or 16 mL) of 0.1 mol/dm^3 NaOH solution is required for complete neutralization. To find the number of moles of NaOH used, we can use the equation:
moles = concentration x volume
moles of NaOH = (0.1 mol/dm^3) x (16 cm^3) / 1000 cm^3/dm^3
moles of NaOH = 0.0016 mol
Step 2: Determine the molar ratio between the acid and NaOH.
From the balanced chemical equation of the neutralization reaction, we know that the ratio of moles between the acid and NaOH is 1:1. This means that 0.0016 mol of NaOH will react with 0.0016 mol of the monobasic acid.
Step 3: Determine the relative mass of the acid.
The relative mass of the acid can be calculated using the equation:
relative mass = mass / moles
Since we are given that the sample of monobasic acid is 0.4 g, we can use the equation to find the relative mass:
relative mass = 0.4 g / 0.0016 mol
relative mass = 250 g/mol
Therefore, the relative mass of the monobasic acid is 250 g/mol.
To determine the relative mass of the monobasic acid, we need to use the equation relating the amount of acid and the amount of base that react during the neutralization process.
From the given information, we know that 16 cm^3 of 0.1 mol/dm^3 NaOH is required to completely neutralize 0.4 g of the acid.
First, let's calculate the number of moles of NaOH used:
- 16 cm^3 is equivalent to 0.016 dm^3 (since 1 dm^3 = 1000 cm^3)
- Concentration of NaOH = 0.1 mol/dm^3
- Moles of NaOH = concentration x volume = 0.1 mol/dm^3 x 0.016 dm^3 = 0.0016 mol
Now, we can use the balanced chemical equation for the neutralization reaction to determine the molar ratio between the acid and the base. Since it is a monobasic acid (meaning it donates one proton or one mole of H+ ions), the balanced equation would be:
Acid + NaOH → Salt + Water
According to the balanced equation, the molar ratio between the acid and NaOH is 1:1.
Therefore, the number of moles of the monobasic acid is also 0.0016 mol.
To calculate the relative mass of the acid, we can use the formula:
Relative mass (in g/mol) = mass (in g) / number of moles
Given that the mass of the acid is 0.4 g and the number of moles is 0.0016 mol, we can substitute these values in the formula:
Relative mass = 0.4 g / 0.0016 mol = 250 g/mol
Therefore, the relative mass of the monobasic acid is 250 g/mol.