What mass of an 85\% pure sample of Mg(OH)_2 is required to...

Moles of $HNO_3 = M \times V = 0.4 \times \dfrac{250}{1000} = 0.1\text{ mol}$

From the balanced chemical equation, $1\text{ mol}$ of $Mg(OH)_2$ neutralizes $2\text{ mol}$ of $HNO_3$.

Moles of $Mg(OH)_2$ required = $\dfrac{0.1}{2} = 0.05\text{ mol}$

Molar mass of $Mg(OH)_2 = 24 + 2 \times (16 + 1) = 58\text{ g/mol}$

Mass of pure $Mg(OH)_2$ required = $0.05 \times 58 = 2.9\text{ g}$

Since the sample is $85\%$ pure, the mass of the sample required is given by:

$m \times \dfrac{85}{100} = 2.9$

$m = \dfrac{2.9 \times 100}{85} = 3.41\text{ g}$

Answer: $3.41\text{ g}$