2.20 g of an ammonium salt was boild with 74 ml of 1 N NaOH till the emission of ammonia gas ceased. The excess of unused NaOH solution required 70 ml of N/2 sulphuric acid for neutralization. The percentage of ammonia in the salt is:

Options

A. 0.31
B. 0.83
C. 0.66
D. 0.47

Answer

To solve the problem step by step, we will follow the outlined procedure to find the percentage of ammonia in the ammonium salt.

### Step 1: Calculate the total equivalents of NaOH used
Given:
- Volume of NaOH = 74 mL = 0.074 L
- Normality of NaOH = 1 N

Using the formula for equivalents:
\[ \text{Equivalents of NaOH} = \text{Normality} \times \text{Volume (L)} = 1 \, \text{N} \times 0.074 \, \text{L} = 0.074 \, \text{equivalents} \]

### Step 2: Calculate the equivalents of H2SO4 used
Given:
- Volume of H2SO4 = 70 mL = 0.070 L
- Normality of H2SO4 = 1/2 N = 0.5 N

Using the formula for equivalents:
\[ \text{Equivalents of H2SO4} = \text{Normality} \times \text{Volume (L)} = 0.5 \, \text{N} \times 0.070 \, \text{L} = 0.035 \, \text{equivalents} \]

### Step 3: Calculate the equivalents of NaOH that reacted with the ammonium salt
The equivalents of NaOH that reacted with the ammonium salt can be calculated by subtracting the equivalents of H2SO4 from the total equivalents of NaOH:
\[ \text{Equivalents of NH4+ (from salt)} = \text{Equivalents of NaOH} - \text{Equivalents of H2SO4} = 0.074 - 0.035 = 0.039 \, \text{equivalents} \]

### Step 4: Calculate the weight of ammonia produced
The equivalent weight of ammonia (NH3) is calculated as follows:
- Molecular weight of NH3 = 17 g/mol
- n-factor of NH3 = 1 (since it can donate one H+ ion)

Thus, the equivalent weight of NH3 is:
\[ \text{Equivalent weight of NH3} = \frac{\text{Molecular weight}}{\text{n-factor}} = \frac{17}{1} = 17 \, \text{g/equivalent} \]

Now, calculate the weight of ammonia produced:
\[ \text{Weight of NH3} = \text{Equivalents of NH3} \times \text{Equivalent weight of NH3} = 0.039 \, \text{equivalents} \times 17 \, \text{g/equivalent} = 0.663 \, \text{g} \]

### Step 5: Calculate the fraction of ammonia in the ammonium salt
Given the total weight of the ammonium salt is 2.20 g, we can find the fraction:
\[ \text{Fraction of NH3} = \frac{\text{Weight of NH3}}{\text{Total weight of salt}} = \frac{0.663}{2.20} \approx 0.3014 \]

### Step 6: Calculate the percentage of ammonia in the ammonium salt
To find the percentage of ammonia in the salt:
\[ \text{Percentage of NH3} = \text{Fraction of NH3} \times 100 = 0.3014 \times 100 \approx 30.14\% \]

### Final Answer
The percentage of ammonia in the ammonium salt is approximately **30.14%**.

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