The specific gravity of concentrated tetraoxosulphate (VI) acid is 1.84. Calculate the volume of tetraoxosulphate (VI) acid required to make 0.01M H2SO4, if the percentage purity of H2SO4 is 98.7.

Molar mass of H$_2$SO$_4$ = $2(\text{1.01 g/mol for H}) + \text{32.07 g/mol for S} + 4(\text{16.00 g/mol for O})$

Density of concentrated H$_2$SO$_4$ = Specific gravity $\times$ Density of water = $1.84 \times 1 \text{ g/mL} = 1.84 \text{ g/mL}$.
Mass of 1 L (1000 mL) of concentrated acid = $\text{Density} \times \text{Volume}$

Mass of pure H$_2$SO$_4$ in 1 L of concentrated acid = $\text{Mass of solution} \times \text{Purity percentage}$

Moles of pure H$_2$SO$_4$ in 1 L of concentrated acid = $\frac{\text{Mass}}{\text{Molar mass}}$

Molarity of concentrated H$_2$SO$_4$ ($\text{M}_1$) = $\text{18.515 M}$.

We use the dilution formula: $\text{M}_1\text{V}_1 = \text{M}_2\text{V}_2$
Where:
$\text{M}_1 = 18.515 \text{ M}$ (molarity of concentrated acid)
$\text{V}_1 = ?$ (volume of concentrated acid required)
$\text{M}_2 = 0.01 \text{ M}$ (target molarity of dilute acid)
$\text{V}_2 = 1000 \text{ mL}$ (target volume of dilute acid, assumed)

Rounding to three significant figures:

The volume of concentrated tetraoxosulphate (VI) acid required is $\boxed{\text{0.540 mL}}$.

Molar mass of $(\text{NH}_4)_2\text{SO}_4$ = $2(\text{14.01 g/