Solutions of hydrogen in palladium may be formed by exposing Pd metal to H2 gas. The concentration of hydrogen in the palladium depends on the pressure of H2 gas applied, but in a more complex fashion than can be described by Henry's law. Under certain conditions, 0.712g of hydrogen gas is dissolved in 252g of palladium metal. (The density of the resulting solution is11.6g/cm3.)

(a)

Determine the molarity of this solution.

*M*

(b)

Determine the molality of this solution.

*m*

(c)

Determine the percent by mass of hydrogen in this solution.

%

To solve this problem, we need to calculate the molarity, molality, and percent by mass of the hydrogen solution in palladium metal.

### (a) Molarity (M)

Molarity is defined as the number of moles of solute (in this case, H₂) per liter of solution.

1. **Find the moles of hydrogen (H₂):**

   - Molar mass of H₂ = 2.02 g/mol
   - Given mass of H₂ = 0.712 g

   \[
   \text{Moles of H₂} = \frac{0.712 \text{ g}}{2.02 \text{ g/mol}} = 0.3525 \text{ mol}
   \]

2. **Calculate the volume of the solution:**

   - Total mass of the solution (H₂ + Pd) = 0.712 g + 252 g = 252.712 g
   - Density of the solution = 11.6 g/cm³

   \[
   \text{Volume of the solution} = \frac{\text{Total mass of the solution}}{\text{Density of the solution}} = \frac{252.712 \text{ g}}{11.6 \text{ g/cm³}} = 21.78 \text{ cm³} = 0.02178 \text{ L}
   \]

3. **Calculate the molarity:**

   \[
   M = \frac{\text{Moles of H₂}}{\text{Volume of the solution in liters}} = \frac{0.3525 \text{ mol}}{0.02178 \text{ L}} \approx 16.18 \text{ M}
   \]

### (b) Molality (m)

Molality is defined as the number of moles of solute per kilogram of solvent.

1. **Moles of H₂** (already calculated) = 0.3525 mol
2. **Mass of palladium (solvent):** 

   - Mass of palladium = 252 g = 0.252 kg

3. **Calculate the molality:**

   \[
   m = \frac{\text{Moles of H₂}}{\text{Mass of solvent in kg}} = \frac{0.3525 \text{ mol}}{0.252 \text{ kg}} \approx 1.40 \text{ m}
   \]

### (c) Percent by mass of hydrogen

Percent by mass is calculated as the mass of the solute divided by the total mass of the solution, multiplied by 100.

\[
\%\text{H₂} = \left(\frac{\text{Mass of H₂}}{\text{Total mass of the solution}}\right) \times 100 = \left(\frac{0.712 \text{ g}}{252.712 \text{ g}}\right) \times 100 \approx 0.282\%
\]

### Summary of Results

- **(a) Molarity (M):** \( \approx 16.18 \, \text{M} \)
- **(b) Molality (m):** \( \approx 1.40 \, \text{m} \)
- **(c) Percent by mass of hydrogen:** \( \approx 0.282\% \)

Would you like further details or have any questions?

Here are 5 related questions:
1. How does molarity change if the density of the solution increases?
2. What is the difference between molality and molarity?
3. How would the calculations change if the solute were another gas, such as oxygen?
4. Why is molality independent of temperature, unlike molarity?
5. How would the percent by mass change if the amount of Pd was increased?

**Tip:** Always remember to distinguish between molarity and molality in calculations, especially in scenarios involving temperature changes.

Learn how to calculate molarity, molality, and percent composition in a solution of hydrogen dissolved in palladium metal. This problem involves solution chemistry concepts such as molarity, molality, and percent composition. Suitable for undergraduate-level students and chemistry enthusiasts.

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