To 40 g of a 12% sulfuric acid solution was added 4 g of sulfur oxide (VI). Calculate the mass fraction of the substance (%) in the new solution.
SO3 + H2O = H2SO4
n (SO3) = 4/80 = 0.05 mol
⇒ n (H2SO4) formed = 0.05 x 98 = 4.9 g
m (H2SO4) ref. solution = 40 x 0.12 = 4.8 g
m (H2O) ref. solution = 40 – 4.8 = 35.2 g
m (H2O for SO3) = 0.12 x 18 = 2.16 g
m (H2SO4 total) = 4.9 + 4.8 = 9.7 g
m (final solution) = 40 + 4 = 44 g
ω (H2SO4 in final solution) = 9.7 / 44 x 100% = 22%
Answer: the mass fraction of sulfuric acid in the final solution is 22%.
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