When 2.0 grams of copper (II) nitrate is added to 1000 ml of pure water, by how much is the vapor pressure of water decreased, given that at 20°C the vapor pressure of pure water is 17.535 mm Hg?

Options

A. 0.303
B. 0.0333
C. 0.0033
D. 3.14 x `10^-4`

Answer

To solve the problem of how much the vapor pressure of water decreases when 2.0 grams of copper (II) nitrate is added to 1000 ml of pure water, we can follow these steps:

### Step 1: Determine the molar mass of copper (II) nitrate
Copper (II) nitrate has the formula Cu(NO₃)₂. The molar mass can be calculated as follows:
- Copper (Cu): 63.55 g/mol
- Nitrogen (N): 14.01 g/mol (there are 2 nitrogen atoms)
- Oxygen (O): 16.00 g/mol (there are 6 oxygen atoms)

Calculating the molar mass:
\[ \text{Molar mass of Cu(NO₃)₂} = 63.55 + (2 \times 14.01) + (6 \times 16.00) = 63.55 + 28.02 + 96.00 = 187.57 \text{ g/mol} \]

### Step 2: Calculate the number of moles of copper (II) nitrate
Using the mass of copper (II) nitrate and its molar mass, we can find the number of moles:
\[ \text{Number of moles of Cu(NO₃)₂} = \frac{\text{mass}}{\text{molar mass}} = \frac{2.0 \text{ g}}{187.57 \text{ g/mol}} \approx 0.01065 \text{ moles} \]

### Step 3: Calculate the number of moles of water
The volume of water is given as 1000 ml, which is equivalent to 1000 g (assuming the density of water is 1 g/ml). The molar mass of water (H₂O) is 18.02 g/mol.
\[ \text{Number of moles of water} = \frac{1000 \text{ g}}{18.02 \text{ g/mol}} \approx 55.51 \text{ moles} \]

### Step 4: Calculate the mole fraction of the solute (copper (II) nitrate)
The mole fraction \( x_2 \) of the solute can be calculated using the formula:
\[ x_2 = \frac{\text{moles of solute}}{\text{moles of solute} + \text{moles of solvent}} = \frac{0.01065}{0.01065 + 55.51} \approx 1.91 \times 10^{-4} \]

### Step 5: Calculate the decrease in vapor pressure
Using Raoult's Law, the decrease in vapor pressure \( \Delta P \) can be calculated as:
\[ \Delta P = P^0_{\text{water}} \times x_2 \]
Where \( P^0_{\text{water}} \) is the vapor pressure of pure water (17.535 mm Hg).
\[ \Delta P = 17.535 \text{ mm Hg} \times 1.91 \times 10^{-4} \approx 0.00335 \text{ mm Hg} \]

### Conclusion
The decrease in vapor pressure of water when 2.0 grams of copper (II) nitrate is added to 1000 ml of pure water is approximately **0.00335 mm Hg**.

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