To prepare \(250\, \ mL\) standard solution of \(M/20\, \ NaCl\), amount of \(NaCl\) required is:
Answer & explanation
Correct answer: option 2
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.