What weight of sodium hydroxide is required to neutralise 100 ml of `"0.1 HCl"`?

Options

A. 4g
B. 0.4g
C. 0.04g
D. 40 g

Explanation

To solve the problem of determining the weight of sodium hydroxide (NaOH) required to neutralize 100 ml of 0.1 M hydrochloric acid (HCl), we can follow these steps:

### Step-by-Step Solution:

1. **Determine the number of moles of HCl:**
- The formula to calculate the number of moles is:
\[ \text{Number of moles} = \text{Molarity} \times \text{Volume (in liters)} \]
- Given that the molarity of HCl is 0.1 M and the volume is 100 ml (which is 0.1 L):
\[ \text{Number of moles of HCl} = 0.1 \, \text{mol/L} \times 0.1 \, \text{L} = 0.01 \, \text{mol} \]

2. **Write the neutralization reaction:**
- The balanced chemical equation for the reaction between NaOH and HCl is:
\[ \text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O} \]
- From the equation, we see that 1 mole of NaOH reacts with 1 mole of HCl.

3. **Determine the number of moles of NaOH required:**
- Since the reaction is 1:1, the number of moles of NaOH required is equal to the number of moles of HCl:
\[ \text{Number of moles of NaOH} = 0.01 \, \text{mol} \]

4. **Calculate the mass of NaOH required:**
- The molar mass of NaOH is approximately 40 g/mol.
- To find the mass, use the formula:
\[ \text{Mass} = \text{Number of moles} \times \text{Molar mass} \]
- Therefore:
\[ \text{Mass of NaOH} = 0.01 \, \text{mol} \times 40 \, \text{g/mol} = 0.4 \, \text{g} \]

### Final Answer:
The weight of sodium hydroxide required to neutralize 100 ml of 0.1 M HCl is **0.4 g**.