The vapour pressure of water is 92 mm at 323 K. 18.1 g of urea are dissolved in 100 g of water. The vapour pressure is reduced by 5 mm. Calculate the molar mass of urea.

Explanation

To solve the problem of calculating the molar mass of urea when it is dissolved in water, we can follow these steps:

### Step 1: Understand the Given Data
- The vapor pressure of pure water (P₀) = 92 mm at 323 K.
- The vapor pressure of the solution (P₁) is reduced by 5 mm, so:
\[ P₁ = P₀ - 5 \text{ mm} = 92 \text{ mm} - 5 \text{ mm} = 87 \text{ mm} \]
- Mass of urea (solute) = 18.1 g
- Mass of water (solvent) = 100 g

### Step 2: Calculate the Change in Vapor Pressure
The change in vapor pressure (ΔP) is:
\[ ΔP = P₀ - P₁ = 5 \text{ mm} \]

### Step 3: Apply Raoult's Law
According to Raoult's Law, the change in vapor pressure can be expressed as:
\[ ΔP = \frac{n_{solute}}{n_{solvent}} \times P₀ \]
Where:
- \( n_{solute} \) = number of moles of solute (urea)
- \( n_{solvent} \) = number of moles of solvent (water)

### Step 4: Calculate Moles of Solvent
First, we need to calculate the number of moles of water:
- Molar mass of water (H₂O) = 18 g/mol
\[ n_{solvent} = \frac{100 \text{ g}}{18 \text{ g/mol}} = 5.56 \text{ moles} \]

### Step 5: Calculate Moles of Solute
Let the molar mass of urea be \( M \) (g/mol). The number of moles of urea can be calculated as:
\[ n_{solute} = \frac{18.1 \text{ g}}{M} \]

### Step 6: Substitute Values into Raoult's Law
Substituting \( n_{solute} \) and \( n_{solvent} \) into Raoult's Law:
\[ 5 \text{ mm} = \frac{\frac{18.1}{M}}{5.56} \times 92 \]

### Step 7: Rearrange to Solve for M
Rearranging the equation:
\[ 5 = \frac{18.1 \times 92}{M \times 5.56} \]
\[ M = \frac{18.1 \times 92}{5 \times 5.56} \]

### Step 8: Calculate M
Now, let's calculate \( M \):
\[ M = \frac{1663.2}{27.8} \approx 59.8 \text{ g/mol} \]

### Conclusion
The molar mass of urea is approximately 59.8 g/mol.