You are provided with HNO3 with 70% purity and density of 1.38 g/cm³. Calculate: (i) the concentration of the concentrated HNO3 (ii) Calculate the volume of the acid to be taken to prepare 2dm³ of 0.8 moldm³ of HNO3. (Mr(HNO3)=98)

The density of the solution is $1.38 \, \text{g/cm}^3$.

$$\text{Mass of solution} = \text{Density} \times \text{Volume}$$ $$\text{Mass of solution} = 1.38 \, \text{g/cm}^3 \times 1000 \, \text{cm}^3 = 1380 \, \text{g}$$

The purity of the $\text{HNO}_3$ is $70\%$.

$$\text{Mass of pure HNO}_3 = 0.70 \times 1380 \, \text{g} = 966 \, \text{g}$$

The molar mass of $\text{HNO}_3$ is $98 \, \text{g/mol}$.

$$\text{Moles of HNO}_3 = \frac{\text{Mass of pure HNO}_3}{\text{Molar mass of HNO}_3}$$ $$\text{Moles of HNO}_3 = \frac{966 \, \text{g}}{98 \, \text{g/mol}} = 9.8571 \, \text{mol}$$

The volume of the solution assumed is $1 \, \text{dm}^3$.

$$\text{Concentration} = \frac{\text{Moles of HNO}_3}{\text{Volume of solution in dm}^3}$$ $$\text{Concentration} = \frac{9.8571 \, \text{mol}}{1 \, \text{dm}^3} = 9.8571 \, \text{mol/dm}^3$$

Rounding to three significant figures:

$$\text{Concentration} = \boxed{\text{9.86 mol/dm}^3}$$

$C_1 = 9.8571 \, \text{mol/dm}^3$ (concentration of concentrated acid from part i)
$V_1 = ?$ (volume of concentrated acid needed)
$C_2 = 0.8 \, \text{mol/dm}^3$ (concentration of diluted acid)
$V_2 = 2 \, \text{dm}^3$ (volume of diluted acid)

$$V_1 = \frac{C_2V_2}{C_1}$$ $$V_1 = \frac{0.8 \, \text{mol/dm}^3 \times 2 \, \text{dm}^3}{9.8571 \, \text{mol/dm}^3}$$ $$V_1 = \frac{1.6}{9.8571} \, \text{dm}^3$$ $$V_1 = 0.1623 \, \text{dm}^3$$

Rounding to three significant figures:

$$V_1 = \boxed{\text{0.162 dm}^3}$$