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 needs to 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 decrease in vapor pressure (ΔP) can be related to the mole fraction of the solute (urea in this case) using Raoult's Law. The relative lowering of vapor pressure (ΔP/P₀) is equal to the mole fraction of the solute (X_solute).

Given that the vapor pressure decreases by 5%, we have:
\[ \frac{\Delta P}{P_0} = 0.05 \]
This means:
\[ X_{solute} = 0.05 \]

### Step 2: Set up the mole fraction equation
The mole fraction of the solute (urea) can be expressed as:
\[ X_{solute} = \frac{n_{solute}}{n_{solute} + n_{solvent}} \]
Where:
- \(n_{solute}\) = moles of urea
- \(n_{solvent}\) = moles of water

### Step 3: Calculate the moles of water
The number of moles of water (n_solvent) can be calculated using its mass and molar mass:
\[ n_{solvent} = \frac{mass_{water}}{molar\ mass_{water}} = \frac{171\ g}{18\ g/mol} = 9.5\ mol \]

### Step 4: Substitute into the mole fraction equation
Now substituting into the mole fraction equation:
\[ 0.05 = \frac{n_{solute}}{n_{solute} + 9.5} \]

### Step 5: Solve for n_solute
Let \(n_{solute} = n_B\). Rearranging gives:
\[ 0.05(n_{solute} + 9.5) = n_{solute} \]
\[ 0.05n_{solute} + 0.475 = n_{solute} \]
\[ 0.475 = n_{solute} - 0.05n_{solute} \]
\[ 0.475 = 0.95n_{solute} \]
\[ n_{solute} = \frac{0.475}{0.95} = 0.5\ mol \]

### Step 6: Calculate the mass of urea
To find the mass of urea (W_B), we use the formula:
\[ W_B = n_{solute} \times molar\ mass_{urea} \]
The molar mass of urea (NH₂CONH₂) is approximately 60 g/mol. Thus:
\[ W_B = 0.5\ mol \times 60\ g/mol = 30\ g \]

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

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