We need the dissociation constant of benzoic acid which is 6.3x10^'5. Then using the dissociation formula, ka = x2 / (Mo - x) where Mo is the initial concentration. x is determined then. percent ionization is computed as (x/Mo)* 100%. This is then the final answer.
Answer:
Molarity of Na₂CO₃ = 0.25M
% mass = 2.69
Explanation:
Molarity means mole of solute in 1L of solution
Molar mass of solute (Na₂CO₃) = 105,98 g/m
Moles = mass / molar mass → 6.73 g / 105.98 g/m = 0.0635 m
Mol/L = [M]
0.0635 mol/0.250L = 0.25M
Density of solution = Solution mass / Solution volume
1 g/ml = Solution mass / 250 mL → Solution mass is 250g
% mass will be:
In 250 g of solution we have 6.73 g of solute
in 100 g of solution we have (100 . 6.73)/250 = 2.69
The kind of gases probably in astronaut suit is nitrous oxide
