35.4 mL of HCl is required for the neutralization of a solution containing 0.275 g of sodium hydroxide. The normality of hydrochloric acid is

Options

A. 0.97 N
B. 0.142 N
C. 0.194 N
D. 0.244 N

Explanation

To find the normality of hydrochloric acid (HCl) required to neutralize sodium hydroxide (NaOH), we can follow these steps:

### Step 1: Calculate the moles of sodium hydroxide (NaOH)
To find the number of moles of NaOH, we use the formula:

\[ \text{Moles of NaOH} = \frac{\text{mass of NaOH}}{\text{molar mass of NaOH}} \]

Given:
- Mass of NaOH = 0.275 g
- Molar mass of NaOH = 40 g/mol

Calculating the moles:

\[ \text{Moles of NaOH} = \frac{0.275 \, \text{g}}{40 \, \text{g/mol}} = 0.006875 \, \text{mol} \]

### Step 2: Determine the moles of hydrochloric acid (HCl)
In the neutralization reaction, one mole of NaOH reacts with one mole of HCl. Therefore, the moles of HCl will be equal to the moles of NaOH:

\[ \text{Moles of HCl} = \text{Moles of NaOH} = 0.006875 \, \text{mol} \]

### Step 3: Convert the volume of HCl from mL to L
The volume of HCl is given as 35.4 mL. We need to convert this to liters:

\[ \text{Volume of HCl in L} = \frac{35.4 \, \text{mL}}{1000} = 0.0354 \, \text{L} \]

### Step 4: Calculate the molarity of HCl
Molarity (M) is calculated using the formula:

\[ \text{Molarity} = \frac{\text{moles of solute}}{\text{volume of solution in L}} \]

Substituting the values we have:

\[ \text{Molarity of HCl} = \frac{0.006875 \, \text{mol}}{0.0354 \, \text{L}} \approx 0.1942 \, \text{M} \]

### Step 5: Calculate the normality of HCl
For HCl, which is a strong acid, the normality (N) is equal to the molarity (M) because it provides one equivalent of H⁺ per mole of HCl:

\[ \text{Normality of HCl} = \text{Molarity of HCl} = 0.1942 \, \text{N} \]

### Final Answer:
The normality of hydrochloric acid is approximately **0.1942 N**.

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