To prepare \(250\, \ mL\) standard solution of \(M/20\, \ NaCl\), amount of \(NaCl\) required is:

0.73 g

The correct answer is option 2. 0.73 g.

To prepare a \(250\, \text{mL}\) standard solution of \(M/20\) \(NaCl\), we need to calculate the amount of \(NaCl\) required.

\(M/20\) means the solution is \(1/20\) of 1 Molar (M), so the molarity is:

\(\text{Molarity} = \frac{1}{20} \text{ M} = 0.05 \text{ M}\)

Molarity (\(M\)) is defined as moles of solute per liter of solution. For \(250\, \text{mL}\) (which is \(0.250\, \text{L}\)):

\(\text{Moles of } NaCl = \text{Molarity} \times \text{Volume (in liters)}\)

\(\text{Moles of } NaCl = 0.05 \text{ M} \times 0.250 \text{ L} = 0.0125 \text{ moles}\)

The molar mass of \(NaCl\) is approximately \(58.44 \text{ g/mol}\).

\(\text{Mass of } NaCl = \text{Moles of } NaCl \times \text{Molar mass}\)

\(\text{Mass of } NaCl = 0.0125 \text{ moles} \times 58.44 \text{ g/mol} = 0.7305 \text{ g}\)

Rounding to two decimal places, the mass of \(NaCl\) required is approximately \(0.73 \text{ g}\).

Therefore, the amount of \(NaCl\) required is 0.73 g.