Question

Calculate the molarity of 500ml of a solution containing 3.0g of $95\%H_{2}S{O}_4$.

Answer 1

Assume that the given percentage is in wt/wt,.(Because density is not given)
$\Rightarrow 3\times 0.95gmof{H}_{2}S{O}_4=2.85gm$
$\Rightarrow$ no of moles = wt/gm mol $wt=\frac{2.85}{98}=0.029mol.$
.029 moles are present in 500 ml of solution.
$\Rightarrow$ 1 lit of sol will have $0.029\times 2$ moles.
$\Rightarrow$ molarity = 0.058 M
Note: Generally the percentage will be given in vol/vol for acids. Since density is not given, we considered percentage to be $\frac{wt}{wt}$.