25.0cm3 of a solution containing 2.0g of NaOH per 250cm3 of a solution of H2SO4 for complete neutralization. Calculate the concentration of the acid solution in (i) mol/dm3 (ii) g/dm3?
(H = 1.0, O = 16.0, Na = 23.0, S = 32.0)?
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To calculate the concentration of the acid solution, we need to determine the amount of substance (in moles) of both NaOH and H2SO4 involved in the neutralization reaction.
First, let's calculate the moles of NaOH:
Molar mass of NaOH:
Na = 23.0 g/mol
O = 16.0 g/mol
H = 1.0 g/mol
Molar mass of NaOH = 23.0 + 16.0 + 1.0 = 40.0 g/mol
Moles of NaOH = mass of NaOH / molar mass of NaOH
Moles of NaOH = 2.0 g / 40.0 g/mol
Moles of NaOH = 0.05 mol
Next, let's determine the moles of H2SO4:
Since there is a 1:1 stoichiometric ratio between NaOH and H2SO4, the moles of H2SO4 will be the same as the moles of NaOH.
Moles of H2SO4 = 0.05 mol
The volume of the solution is given as 250 cm^3, which needs to be converted to dm^3:
Volume = 250 cm^3 = 250 / 1000 dm^3 = 0.25 dm^3
Now we can calculate the concentration of the acid solution:
(i) Concentration in mol/dm^3:
Concentration = Moles of substance / Volume of solution
Concentration = 0.05 mol / 0.25 dm^3
Concentration = 0.2 mol/dm^3
(ii) Concentration in g/dm^3:
To calculate the concentration in g/dm^3, we need to multiply the concentration in mol/dm^3 by the molar mass of H2SO4.
Molar mass of H2SO4:
H = 1.0 g/mol
O = 16.0 g/mol
S = 32.0 g/mol
Molar mass of H2SO4 = 2(1.0) + 4(16.0) + 32.0 = 98.0 g/mol
Concentration in g/dm^3 = Concentration in mol/dm^3 * Molar mass of H2SO4
Concentration in g/dm^3 = 0.2 mol/dm^3 * 98.0 g/mol
Concentration in g/dm^3 = 19.6 g/dm^3
Therefore, the concentration of the acid solution is (i) 0.2 mol/dm^3 and (ii) 19.6 g/dm^3.
To calculate the concentration of the acid solution, we need to determine the number of moles of H2SO4 present in the solution before neutralization.
First, let's calculate the number of moles of NaOH in the solution.
Given:
Mass of NaOH = 2.0g
Molar mass of NaOH = (23.0 + 16.0 + 1.0) g/mol = 40.0 g/mol
Using the formula:
Number of moles = Mass / Molar mass
Number of moles of NaOH = 2.0g / 40.0 g/mol = 0.05 mol
Now, let's calculate the number of moles of H2SO4 required for neutralization.
The balanced chemical equation for the neutralization reaction between NaOH and H2SO4 is:
2NaOH + H2SO4 -> Na2SO4 + 2H2O
From the equation, we see that 2 moles of NaOH react with 1 mole of H2SO4. Therefore, the number of moles of H2SO4 required is twice the number of moles of NaOH.
Number of moles of H2SO4 = 2 * 0.05 mol = 0.10 mol
Next, let's calculate the volume of the acid solution.
Given:
Volume of acid solution = 250 cm3
To convert cm3 to dm3, we divide by 1000.
Volume of acid solution = 250 cm3 / 1000 = 0.25 dm3
Finally, we can calculate the concentration of the acid solution in mol/dm3 (Molarity).
Molarity = Number of moles / Volume of solution in dm3
(i) Molarity of acid solution = 0.10 mol / 0.25 dm3 = 0.40 mol/dm3
To calculate the concentration of the acid solution in g/dm3, we need to know the molar mass of H2SO4.
H2SO4 contains:
2 hydrogen atoms (2 * 1.0 g/mol) = 2.0g
1 sulfur atom (32.0 g/mol) = 32.0g
4 oxygen atoms (4 * 16.0 g/mol) = 64.0g
Molar mass of H2SO4 = 2.0g + 32.0g + 64.0g = 98.0 g/mol
(ii) Concentration of the acid solution in g/dm3 = Molarity * Molar mass
= 0.40 mol/dm3 * 98.0 g/mol = 39.2 g/dm3
Therefore, the concentration of the acid solution is (i) 0.40 mol/dm3 and (ii) 39.2 g/dm3.