What volume of nitric acid solution (ω (HNO3) = 70.00%; ρ = 1.413 g / ml) should be added to 10 g of oleum (ω (SO3) = 30.00%) so that the mass fraction of sulfuric acid becomes 2 times more than the mass fraction of nitric acid in the resulting solution?
10 g of oleum contain m (SO3) = 3.00 g and m (H2SO4) = 7.00 g
Therefore, the amount n (SO3) = 3.00 / 80 = 0.0375 mol
Because SO3 + H2O = H2SO4 the amount of sulfuric acid according to the equation
reaction 0.0375 mol and the mass is equal to m (H2SO4) = 98×0.0375 = 3.675 g
m (H2SO4) total = 7.00 + 3.675 = 10.675 g
let the volume of the nitric acid solution be V ml,
then m (HNO3) = (1.413 • V • 0.7) g
ω (H2SO4) / ω (HNO3) = 2; ω (H2SO4) / ω (HNO3) = 10.675 / (1.413 • V • 0.7)
whence V (HNO3 solution) = 5.4 ml
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.