`10 mL` of `0.2 N HCl` and `30 mL` of `0.1 N HCl` to gether exaclty neutralises `40 mL` of solution of `NaOH`, which is also exactly neutralised by a solution in water of `0.61 g` of an organic acid. What is the equivalent weight of the organic acid?

Options

A. 61
B. 91.5
C. 122
D. 183

Explanation

To find the equivalent weight of the organic acid, we need to follow these steps:

### Step 1: Calculate the milliequivalents of HCl
We have two solutions of HCl:
1. 10 mL of 0.2 N HCl
2. 30 mL of 0.1 N HCl

The formula for milliequivalents (mEq) is:
\[ \text{mEq} = \text{Volume (mL)} \times \text{Normality (N)} \]

Calculating for each solution:
- For 10 mL of 0.2 N HCl:
\[ \text{mEq}_{\text{HCl1}} = 10 \, \text{mL} \times 0.2 \, \text{N} = 2 \, \text{mEq} \]

- For 30 mL of 0.1 N HCl:
\[ \text{mEq}_{\text{HCl2}} = 30 \, \text{mL} \times 0.1 \, \text{N} = 3 \, \text{mEq} \]

Now, add the milliequivalents of both HCl solutions:
\[ \text{Total mEq of HCl} = 2 \, \text{mEq} + 3 \, \text{mEq} = 5 \, \text{mEq} \]

### Step 2: Set up the equation for NaOH
According to the problem, the total milliequivalents of HCl neutralize the NaOH solution:
\[ \text{mEq}_{\text{HCl}} = \text{mEq}_{\text{NaOH}} \]
Thus,
\[ \text{mEq}_{\text{NaOH}} = 5 \, \text{mEq} \]

### Step 3: Relate NaOH to the organic acid
The organic acid also neutralizes the same amount of NaOH:
\[ \text{mEq}_{\text{NaOH}} = \text{mEq}_{\text{Organic Acid}} \]
So,
\[ \text{mEq}_{\text{Organic Acid}} = 5 \, \text{mEq} \]

### Step 4: Calculate the equivalent weight of the organic acid
We know that the milliequivalents of the organic acid can be expressed as:
\[ \text{mEq} = \frac{\text{mass (g)}}{\text{Equivalent weight (g)}} \]
Given that the mass of the organic acid is 0.61 g, we can set up the equation:
\[ 5 = \frac{0.61}{E} \]
Where \(E\) is the equivalent weight of the organic acid.

Rearranging gives us:
\[ E = \frac{0.61}{5} \]
Calculating this gives:
\[ E = 0.61 \times 1000 / 5 = 122 \, \text{g} \]

### Conclusion
The equivalent weight of the organic acid is **122 g**.