# 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?

(10.0*g)/(98.08*g*mol^-1)xx1/(0.100*L)=??*mol*L^-1.
Approx. $1 \cdot m o l \cdot {L}^{-} 1$.
$\text{Molarity"="Moles of solute"/"Volume of solution}$, the which operation gives an answer in $m o l \cdot {L}^{-} 1$.
This question would have more realistic if it proposed that $8 \cdot g$ of $S {O}_{3}$ were dissolved in $80 \cdot m L$ of water to give a specific volume of solution (likely $80 \cdot m L$). As an acidic oxide, sulfur trioxide dissolves in water to give stoichiometric ${H}_{2} S {O}_{4}$.