# 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%)

**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