Mass of `112 cm^(3)` of `NH_(3)` gas at S.T.P. is

Options

A. `0.085 g`
B. `0.850 g`
C. `8.5 g`
D. `80.5 g`

Explanation

To find the mass of 112 cm³ of NH₃ gas at standard temperature and pressure (STP), we can follow these steps:

### Step-by-Step Solution:

1. **Understand STP Conditions**:
- Standard Temperature (T) = 0°C = 273 K
- Standard Pressure (P) = 1 atm

2. **Convert Volume to Liters**:
- Given volume = 112 cm³
- Since 1 cm³ = 1 mL, we have 112 mL.
- To convert to liters, we use the conversion:
\[ 112 \, \text{mL} = \frac{112}{1000} \, \text{L} = 0.112 \, \text{L} \]

3. **Use Ideal Gas Law**:
- The ideal gas equation is given by:
\[ PV = nRT \]
- Rearranging for moles (n):
\[ n = \frac{PV}{RT} \]
- Where:
- P = 1 atm
- V = 0.112 L
- R = 0.0821 L·atm/(K·mol)
- T = 273 K

4. **Calculate Moles (n)**:
- Substitute the values into the equation:
\[ n = \frac{(1 \, \text{atm}) \times (0.112 \, \text{L})}{(0.0821 \, \text{L·atm/(K·mol)}) \times (273 \, \text{K})} \]
- Calculate the denominator:
\[ 0.0821 \times 273 \approx 22.4143 \]
- Now calculate n:
\[ n = \frac{0.112}{22.4143} \approx 0.00499 \, \text{mol} \]

5. **Find the Molar Mass of NH₃**:
- The molar mass of NH₃ (Ammonia) = 14 (N) + 3 × 1 (H) = 17 g/mol

6. **Calculate the Mass**:
- Use the formula:
\[ \text{Mass} = n \times \text{Molar Mass} \]
- Substitute the values:
\[ \text{Mass} = 0.00499 \, \text{mol} \times 17 \, \text{g/mol} \approx 0.08483 \, \text{g} \]

7. **Final Answer**:
- The mass of 112 cm³ of NH₃ gas at STP is approximately **0.085 g**.