Given density = 1.836 g/cm³.

Given density $= 1.836 \, \text{g/cm}^3$.

Given percentage purity $= 98\%$.

Given relative formula mass $= 98$.

Since the moles were calculated for $1000 \, \text{cm}^3$ ($1 \, \text{L}$) of solution, this is the molarity.

The concentration of the acid is $\boxed{\text{18.36 M}}$.

7.b)

$M_1$ (concentration of concentrated acid) $= 18.36 \, \text{M}$ (from 7.a))
$V_1$ (volume of concentrated acid) $= ?$
$M_2$ (desired concentration of diluted acid) $= 2 \, \text{M}$
$V_2$ (desired volume of diluted acid) $= 2000 \, \text{cm}^3$

The volume of concentrated sulphuric(VI) acid needed is $\boxed{\text{217.86 cm}^3}$.

8.a)
The criteria for testing the purity of water include checking its boiling point and freezing point. Pure water boils at $100^\circ\text{C}$ and freezes at $0^\circ\text{C}$ at standard atmospheric pressure. Any deviation from these specific temperatures indicates the presence of impurities. Another criterion is electrical conductivity; pure water has very low conductivity, while impurities increase it.