Weight of `CO_(2)` in a 10 L cylinder at 5 atm and `27 .^(@)C` is (A)200 g (B)224 g (c)44 g (d) 89.3 g

Options

A. 200 g
B. 224 g
C. 44 g
D. 89.3 g

Explanation

To find the weight of CO₂ in a 10 L cylinder at 5 atm and 27°C, we can use the Ideal Gas Law, which is given by the equation:

\[ PV = nRT \]

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

### Step 1: Convert temperature to Kelvin
The temperature is given as 27°C. To convert this to Kelvin, we use the formula:

\[ T(K) = T(°C) + 273.15 \]

So,

\[ T = 27 + 273.15 = 300.15 \, K \]

### Step 2: Use the Ideal Gas Law to find the number of moles (n)
We can rearrange the Ideal Gas Law to solve for \( n \):

\[ n = \frac{PV}{RT} \]

Substituting the values we have:

- \( P = 5 \, atm \)
- \( V = 10 \, L \)
- \( R = 0.0821 \, atm·L/(mol·K) \)
- \( T = 300.15 \, K \)

Now substituting these values into the equation:

\[ n = \frac{(5 \, atm)(10 \, L)}{(0.0821 \, atm·L/(mol·K))(300.15 \, K)} \]

### Step 3: Calculate the number of moles
Calculating the denominator:

\[ 0.0821 \times 300.15 \approx 24.726 \]

Now substituting back into the equation for \( n \):

\[ n = \frac{50}{24.726} \approx 2.02 \, moles \]

### Step 4: Calculate the weight of CO₂
The molar mass of CO₂ is approximately 44 g/mol. To find the weight, we multiply the number of moles by the molar mass:

\[ \text{Weight} = n \times \text{Molar mass} \]

So,

\[ \text{Weight} = 2.02 \, moles \times 44 \, g/mol \approx 88.88 \, g \]

### Step 5: Round to appropriate significant figures
Rounding to two decimal places, we get:

\[ \text{Weight} \approx 89.3 \, g \]

### Conclusion
The weight of CO₂ in the cylinder is approximately 89.3 g, which corresponds to option (D).

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