Question

Find the molarity and the normality of a 15% solution by mass of $H_{2}S{O}_4$.
(density of $H_{2}S{O}_4=1.10$ g/mL)

• A. 1.68 M, 3.36 N
• B. 3.36 M, 1.68 N
• C. 0.84 M, 1.68 N
• D. 1.68 M, 0.84 N

Answer 1

Answer: (A)

(I) Calculation of Molarity

1. 15% by mass means 15 g of solute in 100 g of solution.
2. Moles of sulphuric acid = $\frac{\text{given mass}}{\text{molar mass}}=\frac{15}{98}=0.153$
3. the density of sulphuric acid is 1.10 g/mL $Density=\frac{Mass}{volume}$
4. So, volume is $\frac{mass}{density}=\frac{100}{1.1}=90.9mL$
5. The molarity of the solution$=\frac{Numberofmoles}{volumeofsolution}=\frac{0.153}{90.9}=1.68M$

Hence the molarity of the solution is 1.68 M

(II) Calculation of Normality

1. The normality of a solution is given as $Normality=Molarity\times Basicity$
2. Since the n-factor or basicity for sulphuric acid is 2, the normality is $Normality=1.68\times 2=3.36N$

Hence, the normality of the solution is 3.36 N