Question

Commercially produced sulphuric acid is 96.4% by mass and it's specific gravity is 1.84.calculate the molarity of the sulphuric acid soltn?

Answer 1

I make it approx. `18*mol*L^-1`

Explanation:

We want the quotient....`"moles of solute (moles)"/"volume of solution (L)"`..

And we can work from a `1*mL` solution, i.e. `1*mL-=10^-3*L`, BECAUSE we have been quoted `rho_"density"`...and can equate the volume to the mass and molar quantity with the given percentage.

And so `[H_2SO_4]=((1*mLxx1.84*g*mL^-1xx96.4%)/(98.08*g*mol^-1))/(10^-3*L)=??*mol*L^-1`...

Oleum is conc. sulfuric saturated with `SO_3` (the acid anhydride of sulfuric acid)...and this is quite nasty stuff....