<u>Answer:</u> The mass percent of
in the battery acid is 31.9 %
<u>Explanation:</u>
- To calculate the number of moles for given molarity, we use the equation:
![\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}](https://tex.z-dn.net/?f=%5Ctext%7BMolarity%20of%20the%20solution%7D%3D%5Cfrac%7B%5Ctext%7BMoles%20of%20solute%7D%5Ctimes%201000%7D%7B%5Ctext%7BVolume%20of%20solution%20%28in%20mL%29%7D%7D)
We are given:
Molarity of
solution = ![4.462\times 10^{-3}M](https://tex.z-dn.net/?f=4.462%5Ctimes%2010%5E%7B-3%7DM)
Volume of solution = 35.05 mL
Putting values in above equation, we get:
![4.462\times 10^{-3}M=\frac{\text{Moles of }Ba(OH)_2\times 1000}{35.05mL}\\\\\text{Moles of }Ba(OH)_2=0.00016mol](https://tex.z-dn.net/?f=4.462%5Ctimes%2010%5E%7B-3%7DM%3D%5Cfrac%7B%5Ctext%7BMoles%20of%20%7DBa%28OH%29_2%5Ctimes%201000%7D%7B35.05mL%7D%5C%5C%5C%5C%5Ctext%7BMoles%20of%20%7DBa%28OH%29_2%3D0.00016mol)
- The chemical equation for the reaction of barium hydroxide and sulfuric acid follows the reaction:
![Ba(OH)_2+H_2SO_4\rightarrow BaSO_4+2H_2O](https://tex.z-dn.net/?f=Ba%28OH%29_2%2BH_2SO_4%5Crightarrow%20BaSO_4%2B2H_2O)
By Stoichiometry of the reaction:
1 mole of barium hydroxide reacts with 1 mole of sulfuric acid
So, 0.00016 moles of barium hydroxide will react with =
of sulfuric acid.
The calculated moles of sulfuric acid is present in 10 mL of solution.
To calculate the number of moles in 250 mL of solution, we use unitary method:
In 10 mL, the number of moles of sulfuric acid are 0.00016 moles
So, in 250 mL, the number of moles of sulfuric acid will be = ![\frac{0.00016}{10}\times 250=0.004moles](https://tex.z-dn.net/?f=%5Cfrac%7B0.00016%7D%7B10%7D%5Ctimes%20250%3D0.004moles)
The calculated moles of sulfuric acid are present in 1 mL of solution.
- To calculate the number of moles, we use the equation:
![\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}](https://tex.z-dn.net/?f=%5Ctext%7BNumber%20of%20moles%7D%3D%5Cfrac%7B%5Ctext%7BGiven%20mass%7D%7D%7B%5Ctext%7BMolar%20mass%7D%7D)
Molar mass of sulfuric acid = 98 g/mol
Moles of sulfuric acid = 0.004 moles
Putting values in above equation, we get:
![0.004mol=\frac{\text{Mass of }H_2SO_4}{98g/mol}\\\\\text{Mass of }H_2SO_4=0.398g](https://tex.z-dn.net/?f=0.004mol%3D%5Cfrac%7B%5Ctext%7BMass%20of%20%7DH_2SO_4%7D%7B98g%2Fmol%7D%5C%5C%5C%5C%5Ctext%7BMass%20of%20%7DH_2SO_4%3D0.398g)
- To calculate the percentage composition of sulfuric acid in battery acid, we use the equation:
![\%\text{ composition of }H_2SO_4=\frac{\text{Mass of }H_2SO_4}{\text{Mass of battery acid}}\times 100](https://tex.z-dn.net/?f=%5C%25%5Ctext%7B%20composition%20of%20%7DH_2SO_4%3D%5Cfrac%7B%5Ctext%7BMass%20of%20%7DH_2SO_4%7D%7B%5Ctext%7BMass%20of%20battery%20acid%7D%7D%5Ctimes%20100)
Mass of battery acid = 1.227 g
Mass of sulfuric acid = 0.398 g
Putting values in above equation, we get:
![\%\text{ composition of }H_2SO_4=\frac{0.398g}{1.227g}\times 100=31.9\%](https://tex.z-dn.net/?f=%5C%25%5Ctext%7B%20composition%20of%20%7DH_2SO_4%3D%5Cfrac%7B0.398g%7D%7B1.227g%7D%5Ctimes%20100%3D31.9%5C%25)
Hence, the mass percent sulfuric acid in battery acid is 31.9 %.