# 250 ml of a 5 molar and 100 ml of a 0.8 molar sulfuric acid solution are mixed. The resulting volume was

250 ml of a 5 molar and 100 ml of a 0.8 molar sulfuric acid solution are mixed. The resulting volume was added up to 3 liters. Find the molar concentration of the equivalent of the resulting solution Finding the molar concentration of the equivalent of the resulting solution is carried out according to the scheme:
M (H2SO4) = 98 g / mol;
feq (H2SO4) = 1/2 = 0.5;
ME (H2SO4) = fEq (H2SO4). M (H2SO4) = 0.5. 98 = 49 g / mol.
1) Calculate the total amount of sulfuric acid when mixing solutions, we get:
n (H2SO4) = (5. 0.25) + (0.8. 0.1) = 1.33 mol.
From here
2) Find the mass of sulfuric acid:
m (H2SO4) = n (H2SO4). M (H2SO4) = (1.33. 98) = 130.34 g.
3) Determine the molar concentration of the equivalent of sulfuric acid in a 3-liter solution of it according to the formula:
CH (H2SO4) = mH2SO4) / [ME (H2SO4). V] where
CH (H2SO4) – molar concentration of the equivalent of a solution of sulfuric acid (mol / l); V is the volume of the total acid solution.
CH (H2SO4) = 130.34 / (49.3) = 0.8866 mol / L (mol · L-1).
Answer: CH (H2SO4) = 0.8866 mol / l Remember: The process of learning a person lasts a lifetime. The value of the same knowledge for different people may be different, it is determined by their individual characteristics and needs. Therefore, knowledge is always needed at any age and position.