Question

Calculate molarity of 9.8% (w/w) solution of H2SO4 if the density of the solution is 1.02g/ml?(molar mass of h2so4 is 98 g)

Answer 1

Molecular weight of H2SO4 is 98.079 g-
mol.
1 ml equals 1.84 g. Hence, 96 ml equals
96 x 1.84 = 176.64 g.
Therefore, 100 ml volume contains
176.64 g H2SO4. So 1 liter contains
1776.4 g.
Hence, the molarity is 1776.4/98.079 =
18.11 M.

Answer 2

Answer: The molarity of the solution is 1.02 M.

Explanation:

We are given:

9.8%(w/w) $H_2SO_4$ solution. This that means 9.8 grams of $H_2SO_4$ is present in 100 gram of solution.

To calculate the molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of the solute}}{\text{Volume of the solution (in L)}}$    ....(1)

We are given:

Density of the solution = 1.02 g/mL

Molar mass of  $H_2SO_4$=98 g/mol

Mass of  $H_2SO_4$ = 9.8

Mass of solution = 100 grams

• To calculate the volume of the solution, we use the equation:

$\text{Density}=\frac{Mass}{Volume}$

$1.02g/mL=\frac{100g}{\text{Volume of the solution}}\\\\\text{Volume of the solution}=98.03mL=0.098L$

• To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

$\text{Number of moles of solution}=\frac{9.8g}{98g/mol}=0.1mol$

Now, putting values in equation 1, we get:

$\text{Molarity of the solution}=\frac{0.1mol}{0.098L}=1.02mol/L$

Hence, the molarity of the solution is 1.02 M.