20.0cm3of a solution containing 4g/L of sodium hydroxide was neutralize by 8cm of dilute sulphuric acid

Question

Calculate the concentration of sulphuric 6 acid in M/L

Answer 1

First, let's calculate the amount (in moles) of sodium hydroxide present in the 20.0 cm^3 solution.

The concentration of sodium hydroxide is given as 4 g/L, which means that for 1 L of solution, there are 4 g of sodium hydroxide.

Since 1 L of solution is equivalent to 1000 cm^3, we can calculate the amount (in moles) of sodium hydroxide in 20.0 cm^3 as follows:

moles of sodium hydroxide = (20.0 cm^3 / 1000 cm^3) * (4 g/L / 40.0 g/mol)
moles of sodium hydroxide = 0.001 mol

Next, let's calculate the concentration of sulfuric acid in mol/L.

From the neutralization reaction between sodium hydroxide (NaOH) and sulfuric acid (H2SO4), we can determine the stoichiometry:

NaOH + H2SO4 -> Na2SO4 + 2H2O

We can see that one mole of sulfuric acid reacts with 2 moles of sodium hydroxide.

Since we have 0.001 moles of sodium hydroxide, we know that 0.001/2 = 0.0005 moles of sulfuric acid were present in the 8 cm^3 of dilute sulfuric acid.

Finally, let's calculate the concentration of sulfuric acid in mol/L.

The volume of the sulfuric acid solution used was 8 cm^3 or 0.008 L.

So, the concentration of sulfuric acid in mol/L is:

Concentration of sulfuric acid = 0.0005 moles / 0.008 L
Concentration of sulfuric acid = 0.0625 mol/L

Therefore, the concentration of the dilute sulfuric acid solution is 0.0625 M/L.