Question

If 50 mL of sulphuric acid yields 6 g of barium sulfate when mixed with excess barium chloride, what is the concentration of sulphuric acid in mol/L?

Answer 1

$m=\dfrac{98g/mol*6g}{233.38g/mol}=2.519g$

$c(H2SO4)=/dfrac{2.519g}{98g/mol*0.05L}=0.514g/mol$