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A 1.19 g sample of an unknown halogen occupies 109 mL at 398 K and 1.41 atm. What is the identity of the halogen?

(Hint: Use the ideal gas law to find [tex]n[/tex] (moles). Then divide mass by moles to find molar mass. Then match this molar mass to the molar mass of the halogens: [tex]F_2[/tex], [tex]Cl_2[/tex], [tex]Br_2[/tex], and [tex]I_2[/tex].)

1. Bromine
2. Fluorine
3. Chlorine
4. Iodine
5. Germanium

Sagot :

GinnyAnsweridn
To identify the unknown halogen, we can follow a step-by-step process using the given data and the ideal gas law. Here's the solution:

### Step 1: Convert Volume to Liters
The volume provided is in milliliters (mL), and we need it in liters (L) for the ideal gas law calculation.

[tex]\[ \text{Volume} = 109 \, \text{mL} \][/tex]
[tex]\[ \text{Volume (L)} = 109 \, \text{mL} \times \frac{1 \, \text{L}}{1000 \, \text{mL}} = 0.109 \, \text{L} \][/tex]

### Step 2: Use Ideal Gas Law to Find Moles ([tex]\( n \)[/tex])

The ideal gas law is given by:

[tex]\[ PV = nRT \][/tex]

Where:
- [tex]\( P \)[/tex] is the pressure (1.41 atm)
- [tex]\( V \)[/tex] is the volume in liters (0.109 L)
- [tex]\( n \)[/tex] is the number of moles of gas
- [tex]\( R \)[/tex] is the universal gas constant (0.0821 L atm / K mol)
- [tex]\( T \)[/tex] is the temperature in Kelvin (398 K)

Rearranging the formula to solve for [tex]\( n \)[/tex]:

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

Substituting in the values:

[tex]\[ n = \frac{(1.41 \, \text{atm}) (0.109 \, \text{L})}{(0.0821 \, \text{L atm / K mol}) (398 \, \text{K})} \][/tex]

[tex]\[ n = \frac{0.15369 \, \text{atm L}}{32.6558 \, \text{L atm / K mol}} \][/tex]

[tex]\[ n = 0.004703 \, \text{mol} \][/tex]

### Step 3: Calculate the Molar Mass

The molar mass ([tex]\( M \)[/tex]) is found by dividing the mass of the sample by the number of moles:

[tex]\[ M = \frac{\text{mass}}{\text{moles}} \][/tex]

Given:
- Mass of the sample = 1.19 g
- Number of moles = 0.004703 mol

[tex]\[ M = \frac{1.19 \, \text{g}}{0.004703 \, \text{mol}} \][/tex]

[tex]\[ M = 253.004 \, \text{g/mol} \][/tex]

### Step 4: Identify the Halogen

We compare the calculated molar mass with the known molar masses of the halogens:

- Fluorine (F[tex]\(_2\)[/tex]): 37.9968064 g/mol
- Chlorine (Cl[tex]\(_2\)[/tex]): 70.906 g/mol
- Bromine (Br[tex]\(_2\)[/tex]): 159.808 g/mol
- Iodine (I[tex]\(_2\)[/tex]): 253.80894 g/mol

The calculated molar mass (253.004 g/mol) is closest to the molar mass of iodine (253.80894 g/mol).

Thus, the identity of the halogen is:

[tex]\[ \boxed{\text{Iodine}} \][/tex]
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