Question

$H_{2}S{O}_4$ solution is $98\%$ by weight of solution with density $1.8gm{L}^{-1}$. Express concentration as percent $H_{2}S{O}_4$ weight by volume of solution.

Answer 1

• $98\%$ by weight of solution means $98g$ of $H_{2}S{O}_4$ is present in $100g$ of solution.
  
• Volume of solution $=\frac{\text{mass of solution}}{\text{density of solution}}=\frac{100g}{(1.8\times 1000)g/L}$
• Volume of solution $=\frac{1}{18}L$
• Per cent $H_{2}S{O}_4\text{w/v}$ of solution $=$ molarity $=\frac{\text{number of moles of}{H}_{2}S{O}_4}{\text{volume of solution (in L)}}$
• Molarity $=\frac{\text{given mass of}{H}_{2}S{O}_4/\text{molar mass of}{H}_{2}S{O}_4}{\text{volume of solution}}$
• Molarity $=\frac{98/98}{1/18}=18M$