0.6 g of urea on strong heating with NaOH evolves NH3. Liberated NH3 will combine completely with which of the following HCl solution ?
A
100ml of 0.2N HCl
B
400ml of 0.2N HCl
C
100ml of 0.1N HCl
D
200ml of 0.2N HCl
Text Solution
AI Generated Solution
The correct Answer is:
To solve the problem, we need to follow these steps:
### Step 1: Determine the molecular mass of urea.
Urea has the formula \( \text{CO(NH}_2\text{)}_2 \). The molecular mass can be calculated as follows:
- Carbon (C): 12 g/mol
- Oxygen (O): 16 g/mol
- Nitrogen (N): 14 g/mol (2 Nitrogens)
- Hydrogen (H): 1 g/mol (4 Hydrogens)
Calculating the total:
\[
\text{Molecular mass of urea} = 12 + 16 + (2 \times 14) + (4 \times 1) = 12 + 16 + 28 + 4 = 60 \text{ g/mol}
\]
### Step 2: Calculate the number of moles of urea.
Using the formula for moles:
\[
\text{Number of moles} = \frac{\text{Given mass}}{\text{Molecular mass}} = \frac{0.6 \text{ g}}{60 \text{ g/mol}} = 0.01 \text{ moles}
\]
### Step 3: Determine the moles of ammonia produced.
From the reaction of urea with NaOH, we know that 1 mole of urea produces 2 moles of ammonia. Therefore:
\[
\text{Moles of NH}_3 = 0.01 \text{ moles of urea} \times 2 = 0.02 \text{ moles of NH}_3
\]
### Step 4: Determine the moles of HCl required for neutralization.
Since ammonia (NH₃) is a base and HCl is an acid, they will react in a 1:1 ratio:
\[
\text{Moles of HCl required} = \text{Moles of NH}_3 = 0.02 \text{ moles}
\]
### Step 5: Analyze the given HCl solutions to find the correct one.
We need to find which HCl solution can provide 0.02 moles of HCl. The formula to calculate moles from concentration is:
\[
\text{Moles} = \text{Normality} \times \text{Volume (L)}
\]
If the volume is given in mL, we convert it to liters by dividing by 1000.
### Step 6: Check each option.
1. **Option 1:** \(0.2 \text{ N} \times \frac{100 \text{ mL}}{1000} = 0.2 \times 0.1 = 0.02 \text{ moles} \) (Correct)
2. **Option 2:** \(0.2 \text{ N} \times \frac{400 \text{ mL}}{1000} = 0.2 \times 0.4 = 0.08 \text{ moles} \) (Incorrect)
3. **Option 3:** \(0.1 \text{ N} \times \frac{100 \text{ mL}}{1000} = 0.1 \times 0.1 = 0.01 \text{ moles} \) (Incorrect)
4. **Option 4:** \(0.2 \text{ N} \times \frac{200 \text{ mL}}{1000} = 0.2 \times 0.2 = 0.04 \text{ moles} \) (Incorrect)
### Conclusion:
The correct answer is **Option 1**, which provides exactly 0.02 moles of HCl.
