[Sis290025]

1) In the preparation of a solution of sulfuric acid, 26.7g of SO3 was dissolved in enough water to prepare 8.20L of solution. What is the concentration of sulfur (moles per liter) in this solution?

Equation given: SO3 (g) + H2O (l) --> H2SO4 (aq), which is already balanced

M = mol of solute/ L of solution

I don't know if this is correct because I am confused about finding the concentration of just sulfur:

26.7 g SO3 (1 mol SO3 / 80.07 g SO3)* (1 mol H2SO4/1 mol HSO3) = 0.33346 mol H2SO4

Now is it (0.33326 mol SO4 )/ 8.20 L = M = 0.406659? but this is the M of H2SO4 not S?

To get M of S, do I multiply the mol of SO4 by sulfur's mass composition like so:
 0.33346  mol H2SO4 *(.32687) = .109 mol of S or is this completely wrong? Is the M for H2SO4 same as M for sulfur?

Please reply soon. Thanks.

[mike]

mass of SO3 = 26.7g
molecular weight of SO3 = 80g.mol-1
n = 0.33 moles

0.33 moles is the number of moles of SO3, which according to your equation is equal to the number of moles of H2SO4 (or SO4).

now, 0.33 moles of H2SO4 contains 0.33 moles of S (0.66 moles of H and 1.32 moles O),

so: number of moles of S in 0.33 moles, and volume is 8.20L therefore concentration is:

c = n/v