Calculate the molarity of a $98.52\%$ solution of sulfuric acid ($d$ $=$ $1.8405$).
Hint: The molarity of a solution does not depend on the volume collected. So this volume may be fixed freely!
Let's take $1L =1000 cm^3$ of the solution which we will call $S$:
$\rho_S$ $=$ $1.8405\frac{g}{cm^3}$
1) $m_S$ $=$ $\rho_S\cdot V_S$ $=$ $1.8405\cdot 1000$ $=$ $1840.5g$
2) $m_{H_2SO_4}$ $=$ $\frac{m_S\cdot \%_{H_2SO_4}}{100}$ $=$ $\frac{1840.5\cdot 98.52}{100}$ $=$ $1813g$
3) $n_{H_2SO_4}$ $=$ $\frac{m_{H_2SO_4}}{M_{H_2SO_4}}$ $=$ $\frac{1813}{98}$ $=$ $18.50 mol$
4) $[H_2SO_4]$ $=$ $\frac{n_{H_2SO_4}}{V_S}$ $=$ $\frac{18.50}{1}$ $=$ $18.50\frac{mol}{L}$
