A certain hydrate has the formula `"MgSO"_(4).xH_(2)O`. A quantity of 54.2 g of the compound is heated in an oven to drive off the water. If the steam generated exerts a pressure of 24.8 atm in a 2.0 L container at `120^(@)C`, calculate `x`.

Options

A. 2
B. 5
C. 6
D. 7

Explanation

To solve the problem, we will follow these steps:

### Step 1: Use the Ideal Gas Law to find the number of moles of water vapor (steam).
The Ideal Gas Law is given by the equation:

\[ PV = nRT \]

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

Given:
- \( P = 24.8 \, \text{atm} \)
- \( V = 2.0 \, \text{L} \)
- \( T = 120^\circ C = 120 + 273 = 393 \, \text{K} \)

Now, rearranging the Ideal Gas Law to solve for \( n \):

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

Substituting the values:

\[ n = \frac{(24.8 \, \text{atm}) \times (2.0 \, \text{L})}{(0.0821 \, \text{L·atm/(K·mol)}) \times (393 \, \text{K})} \]

Calculating \( n \):

\[ n = \frac{49.6}{32.2773} \approx 1.537 \, \text{moles of } H_2O \]

### Step 2: Calculate the mass of water.
To find the mass of water, we use the molar mass of water, which is approximately 18 g/mol.

\[ \text{Mass of } H_2O = n \times \text{molar mass} \]

\[ \text{Mass of } H_2O = 1.537 \, \text{moles} \times 18 \, \text{g/mol} \]

Calculating the mass:

\[ \text{Mass of } H_2O \approx 27.69 \, \text{g} \]

### Step 3: Calculate the mass of MgSO4.
The total mass of the hydrate is given as 54.2 g. To find the mass of MgSO4, we subtract the mass of water from the total mass.

\[ \text{Mass of } MgSO_4 = \text{Total mass} - \text{Mass of } H_2O \]

\[ \text{Mass of } MgSO_4 = 54.2 \, \text{g} - 27.69 \, \text{g} \]

Calculating the mass:

\[ \text{Mass of } MgSO_4 \approx 26.51 \, \text{g} \]

### Step 4: Calculate the number of moles of MgSO4.
The molar mass of MgSO4 is approximately 120 g/mol.

\[ n_{MgSO_4} = \frac{\text{Mass of } MgSO_4}{\text{Molar mass of } MgSO_4} \]

\[ n_{MgSO_4} = \frac{26.51 \, \text{g}}{120 \, \text{g/mol}} \]

Calculating the number of moles:

\[ n_{MgSO_4} \approx 0.2209 \, \text{moles} \]

### Step 5: Calculate the value of x.
The value of x in the hydrate formula \( MgSO_4 \cdot xH_2O \) can be found using the ratio of moles of water to moles of MgSO4.

\[ x = \frac{n_{H_2O}}{n_{MgSO_4}} \]

\[ x = \frac{1.537 \, \text{moles}}{0.2209 \, \text{moles}} \]

Calculating x:

\[ x \approx 6.96 \approx 7 \]

Thus, the value of x is 7.

### Final Answer:
The formula of the hydrate is \( MgSO_4 \cdot 7H_2O \).

---