Question

What is the molarity of a solution formed by mixing 10.0 g of `H_2SO_4` with enough water to make 0.100 L of solution?

Answer 1

`(10.0*g)/(98.08*g*mol^-1)xx1/(0.100*L)=??*mol*L^-1`.

Approx. `1*mol*L^-1`.

Explanation:

`"Molarity"="Moles of solute"/"Volume of solution"`, the which operation gives an answer in `mol*L^-1`.

This question would have more realistic if it proposed that `8*g` of `SO_3` were dissolved in `80*mL` of water to give a specific volume of solution (likely `80*mL`). As an acidic oxide, sulfur trioxide dissolves in water to give stoichiometric `H_2SO_4`.