What volume of a solution of hydrochloric acid containing `73 g` acid per litre would sufficient for the exact neutralisation of sodium hydroxide obtained by allowing `0.46 g` of metallic sodium to act upon water.

Explanation

To solve the problem, we need to determine the volume of hydrochloric acid (HCl) solution required to neutralize the sodium hydroxide (NaOH) produced from the reaction of metallic sodium with water. Here’s the step-by-step solution:

### Step 1: Calculate the moles of sodium (Na)
The first step is to find out how many moles of sodium we have. The mass of sodium given is 0.46 g, and the molar mass of sodium (Na) is 23 g/mol.

\[ \text{Moles of Na} = \frac{\text{mass of Na}}{\text{molar mass of Na}} = \frac{0.46 \, \text{g}}{23 \, \text{g/mol}} = 0.02 \, \text{moles} \]

### Step 2: Determine the moles of sodium hydroxide (NaOH) produced
From the reaction of sodium with water, we know that 2 moles of sodium produce 2 moles of NaOH. Therefore, the moles of NaOH produced will be equal to the moles of Na used.

\[ \text{Moles of NaOH} = 0.02 \, \text{moles} \]

### Step 3: Establish the relationship between moles of HCl and NaOH
For complete neutralization, the moles of hydrochloric acid (HCl) required will be equal to the moles of sodium hydroxide (NaOH).

\[ \text{Moles of HCl} = \text{Moles of NaOH} = 0.02 \, \text{moles} \]

### Step 4: Calculate the molarity of the HCl solution
The problem states that the hydrochloric acid solution contains 73 g of HCl per liter. To find the molarity, we need to calculate the number of moles of HCl in 73 g.

The molar mass of HCl is approximately 36.5 g/mol.

\[ \text{Moles of HCl in 1 L} = \frac{73 \, \text{g}}{36.5 \, \text{g/mol}} \approx 2 \, \text{moles} \]

Thus, the molarity (M) of the HCl solution is:

\[ \text{Molarity of HCl} = 2 \, \text{moles/L} \]

### Step 5: Use the molarity to find the volume of HCl required
Using the relationship between moles, molarity, and volume, we can find the volume of HCl needed for neutralization.

\[ \text{Moles of HCl} = \text{Molarity of HCl} \times \text{Volume of HCl} \]

Rearranging gives us:

\[ \text{Volume of HCl} = \frac{\text{Moles of HCl}}{\text{Molarity of HCl}} = \frac{0.02 \, \text{moles}}{2 \, \text{moles/L}} = 0.01 \, \text{L} = 10 \, \text{mL} \]

### Final Answer
The volume of hydrochloric acid solution required for the exact neutralization of sodium hydroxide is **10 mL**.

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