Question

What is the molarity of $H_{2}S{O}_4$ solution that has a density of $1.84{\text{g cc}}^{-1}$ and contains 98% by mass of $H_{2}S{O}_4$? (Given atomic mass of S = 32 amu)

• A. $4.18M$
• B. $8.14M$
• C. $18.4M$
• D. $18M$

Answer 1

The correct option is C $18.4M$

Given:
Mass% = 98% i.e. 98 g of $H_{2}S{O}_4$ is present in 100 g of the solution.
Density $=1.84{\text{g cc}}^{-1}=1.84{\text{g mL}}^{-1}$

Hence, volume $=\frac{100}{1.84}\text{mL}=\frac{1}{18.4}\text{L}$
Given mass of $H_{2}S{O}_4=98g$
Number of moles of $H_{2}S{O}_4=\frac{98}{98}=1$

$\text{Molarity}=\frac{\text{moles}}{\text{volume of the solution in L}}$

$\text{Molarity}=\frac{1}{1/18.4}=18.4M$