100 g of HCI solution with relative density 1.117 g/ml contains 33.4 g of HCI. What volume of this HCI solution will be required to exactly neutralize 5 litre of N/10 NaOH solution?

Options

A. 77.8 ml
B. 48.9 ml
C. 23.2 ml
D. 86.2 ml

Answer

To solve the problem step-by-step, we will follow the given information and apply the concepts of chemistry related to solutions, normality, and neutralization reactions.

### Step 1: Determine the Density and Volume of HCl Solution
Given:
- Mass of HCl solution = 100 g
- Relative density (RD) = 1.117 g/ml

First, we need to find the volume of the HCl solution using the formula:
\[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \]
Rearranging gives us:
\[ \text{Volume} = \frac{\text{Mass}}{\text{Density}} \]
Substituting the values:
\[ \text{Volume} = \frac{100 \text{ g}}{1.117 \text{ g/ml}} \approx 89.52 \text{ ml} \]

### Step 2: Calculate the Gram Equivalent of HCl in the Solution
Given:
- Mass of HCl in the solution = 33.4 g
- Molar mass of HCl = 36.5 g/mol

The equivalent mass of HCl is equal to its molar mass since the n-factor (number of H+ ions produced) for HCl is 1. Therefore:
\[ \text{Gram Equivalent} = \frac{\text{Mass of HCl}}{\text{Equivalent mass}} = \frac{33.4 \text{ g}}{36.5 \text{ g/mol}} \approx 0.915 \text{ equivalents} \]

### Step 3: Calculate the Normality of HCl Solution
Normality (N) is given by:
\[ N = \frac{\text{Gram Equivalent}}{\text{Volume in L}} \]
First, convert the volume from ml to L:
\[ \text{Volume in L} = \frac{89.52 \text{ ml}}{1000} = 0.08952 \text{ L} \]
Now, substituting the values:
\[ N = \frac{0.915 \text{ equivalents}}{0.08952 \text{ L}} \approx 10.22 \text{ N} \]

### Step 4: Use Neutralization Reaction to Find Volume of HCl Required
We need to neutralize 5 L of N/10 NaOH solution. The normality of NaOH (N2) is:
\[ N_2 = \frac{1}{10} \text{ N} \]
Using the neutralization equation:
\[ N_1 \cdot V_1 = N_2 \cdot V_2 \]
Where:
- \(N_1 = 10.22 \text{ N}\) (Normality of HCl)
- \(V_1\) = Volume of HCl solution (unknown)
- \(N_2 = 0.1 \text{ N}\) (Normality of NaOH)
- \(V_2 = 5 \text{ L}\)

Substituting the known values:
\[ 10.22 \cdot V_1 = 0.1 \cdot 5 \]
\[ 10.22 \cdot V_1 = 0.5 \]
Now, solving for \(V_1\):
\[ V_1 = \frac{0.5}{10.22} \approx 0.04892 \text{ L} = 48.92 \text{ ml} \]

### Final Answer
The volume of HCl solution required to neutralize 5 L of N/10 NaOH solution is approximately **48.92 ml**.