What mass of urea be dissolved in `171 g` of water so as to decrease the vapour pressure of water by `5%`?

Options

A. `15 g`
B. `20 g`
C. `25 g`
D. `30 g`

Explanation

To solve the problem of finding the mass of urea that must be dissolved in 171 g of water to decrease the vapor pressure of water by 5%, we can follow these steps:

### Step 1: Understand the relationship between vapor pressure lowering and mole fraction
The lowering of vapor pressure (ΔP) is given as a percentage of the original vapor pressure (P₀). In this case, the decrease is 5%, which can be expressed as:
\[ \frac{\Delta P}{P_0} = 0.05 \]
This lowering of vapor pressure is directly proportional to the mole fraction of the solute (urea in this case).

### Step 2: Set up the mole fraction equation
Let \( N_A \) be the number of moles of water (solvent) and \( N_B \) be the number of moles of urea (solute). The mole fraction of the solute (urea) can be expressed as:
\[ X_B = \frac{N_B}{N_A + N_B} \]
Since we know that \( \Delta P/P_0 = 0.05 \), we can write:
\[ X_B = 0.05 \]

### Step 3: Calculate moles of water (solvent)
The number of moles of water can be calculated using its mass and molar mass. The molar mass of water (H₂O) is approximately 18 g/mol. Therefore, the number of moles of water is:
\[ N_A = \frac{171 \text{ g}}{18 \text{ g/mol}} = 9.5 \text{ moles} \]

### Step 4: Substitute into the mole fraction equation
Now we can substitute \( N_A \) into the mole fraction equation:
\[ 0.05 = \frac{N_B}{9.5 + N_B} \]

### Step 5: Solve for \( N_B \)
Cross-multiplying gives:
\[ 0.05(9.5 + N_B) = N_B \]
Expanding this:
\[ 0.475 + 0.05N_B = N_B \]
Rearranging terms:
\[ 0.475 = N_B - 0.05N_B \]
\[ 0.475 = 0.95N_B \]
Now, solving for \( N_B \):
\[ N_B = \frac{0.475}{0.95} \approx 0.5 \text{ moles} \]

### Step 6: Calculate the mass of urea
Now that we have the number of moles of urea, we can find the mass of urea using its molar mass. The molar mass of urea (NH₂CONH₂) is approximately 60 g/mol. Therefore, the mass of urea (W_B) is:
\[ W_B = N_B \times \text{molar mass of urea} = 0.5 \text{ moles} \times 60 \text{ g/mol} = 30 \text{ g} \]

### Final Answer
The mass of urea that must be dissolved in 171 g of water to decrease the vapor pressure by 5% is **30 g**.

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