The volume of 1g each of methane `( CH_(4))` ethane `( C_(2) H_(6))` , propane `( C_(3) H_(8))` and butane `( C_(4) H_(10))` was measured at 350 K and 1 atm. What is the volume of butane ?

Options

A. `495cm^(3)`
B. `600 cm^(3)`
C. ` 900 cm^(3)`
D. ` 1700 cm^(3)`

Answer

To find the volume of butane (C₄H₁₀) at 350 K and 1 atm, we can use the ideal gas law, which is given by the formula:

\[ PV = nRT \]

Where:
- \( P \) = pressure (in atm)
- \( V \) = volume (in liters)
- \( n \) = number of moles of the gas
- \( R \) = universal gas constant (0.0821 L·atm/(K·mol))
- \( T \) = temperature (in Kelvin)

### Step 1: Calculate the molar mass of butane (C₄H₁₀)
The molar mass of butane can be calculated as follows:
- Carbon (C) has a molar mass of approximately 12 g/mol.
- Hydrogen (H) has a molar mass of approximately 1 g/mol.

Thus, the molar mass of butane is:
\[ \text{Molar mass of C₄H₁₀} = (4 \times 12) + (10 \times 1) = 48 + 10 = 58 \text{ g/mol} \]

### Step 2: Calculate the number of moles of butane
Given that we have 1 gram of butane, we can calculate the number of moles (\( n \)) using the formula:
\[ n = \frac{\text{mass}}{\text{molar mass}} \]

Substituting the values:
\[ n = \frac{1 \text{ g}}{58 \text{ g/mol}} = \frac{1}{58} \text{ mol} \]

### Step 3: Substitute values into the ideal gas law
We know:
- \( P = 1 \text{ atm} \)
- \( n = \frac{1}{58} \text{ mol} \)
- \( R = 0.0821 \text{ L·atm/(K·mol)} \)
- \( T = 350 \text{ K} \)

Now, substituting these values into the ideal gas law:
\[ 1 \text{ atm} \cdot V = \left(\frac{1}{58} \text{ mol}\right) \cdot (0.0821 \text{ L·atm/(K·mol)}) \cdot (350 \text{ K}) \]

### Step 4: Solve for volume (V)
Calculating the right-hand side:
\[ V = \frac{\left(\frac{1}{58}\right) \cdot (0.0821) \cdot (350)}{1} \]

Calculating the multiplication:
\[ V = \frac{0.0821 \cdot 350}{58} \]

Calculating \( 0.0821 \cdot 350 \):
\[ 0.0821 \cdot 350 = 28.735 \]

Now divide by 58:
\[ V = \frac{28.735}{58} \approx 0.495 \text{ L} \]

### Step 5: Convert volume from liters to milliliters
Since \( 1 \text{ L} = 1000 \text{ mL} \):
\[ V = 0.495 \text{ L} \times 1000 \text{ mL/L} = 495 \text{ mL} \]

### Conclusion
The volume of butane (C₄H₁₀) at 350 K and 1 atm is approximately **495 mL** or **495 cm³**.

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