Final answer:
The normality of the caustic soda solution with 5 g of NaOH per litre is approximately 0.125N, as calculated by dividing the mass of NaOH by its equivalent weight (which is the same as its molar mass for NaOH) and the volume of the solution.
Step-by-step explanation:
The question asks to calculate the normality of a caustic soda (NaOH) solution containing 5 g of NaOH per litre. Normality (N) is the number of equivalents of solute per liter of solution. The molar mass of NaOH is approximately 40 g/mol, and since NaOH provides 1 equivalent of OH- per molecule (as it is a monoprotic base), it has an equivalence factor of 1.
To find the normality (N), you use the formula:
N = (Mass of solute in grams) / (Equivalent weight of solute × Volume of solution in liters)
Equivalent weight of NaOH = Molar mass / Equivalence factor = 40 g/mol / 1 = 40 g/equiv.
Substituting the given values:
N = 5 g / (40 g/equiv × 1 L) = 0.125 equiv/L
Therefore, the normality of the solution is approximately 0.125N, which corresponds to option (a).