Pure benzenen boils at `80^(@)C`. If latent heat of vaporization of benzenen is `90 "cal"` per `g`, calculate the molecular weight of solute.

Explanation

To solve the problem, we need to calculate the molecular weight of the solute using the given information about benzene's boiling point and its latent heat of vaporization. Here’s a step-by-step solution:

### Step 1: Convert the boiling point of benzene to Kelvin
The boiling point of pure benzene is given as \(80^\circ C\). To convert this to Kelvin, we use the formula:
\[ T(K) = T(°C) + 273 \]
Substituting the value:
\[ T(K) = 80 + 273 = 353 \, K \]

**Hint**: Remember to always convert Celsius to Kelvin when dealing with temperature in thermodynamic equations.

### Step 2: Identify the latent heat of vaporization
The latent heat of vaporization of benzene is given as \(90 \, \text{cal/g}\).

**Hint**: Latent heat is the amount of heat required to convert a unit mass of a substance from liquid to vapor without a change in temperature.

### Step 3: Calculate the ebullioscopic constant \(K_b\)
The formula for the ebullioscopic constant \(K_b\) is:
\[ K_b = \frac{R T^2}{1000 L} \]
Where:
- \(R\) = 2 (the gas constant in appropriate units)
- \(T\) = 353 K (the boiling point in Kelvin)
- \(L\) = 90 cal/g (latent heat of vaporization)

Substituting the values:
\[ K_b = \frac{2 \times (353)^2}{1000 \times 90} \]
Calculating \(K_b\):
\[ K_b = \frac{2 \times 124609}{90000} = \frac{249218}{90000} \approx 2.77 \, \text{K kg/mol} \]

**Hint**: Ensure you use consistent units when calculating \(K_b\).

### Step 4: Calculate the change in boiling point \(\Delta T\)
The change in boiling point \(\Delta T\) can be calculated as:
\[ \Delta T = T_{boiling} - T_{pure} \]
Given that the boiling point of pure benzene is \(80.175^\circ C\):
\[ \Delta T = 80.175 - 80 = 0.175 \, K \]

**Hint**: \(\Delta T\) is the difference between the boiling point of the solution and the pure solvent.

### Step 5: Use the formula to find the molecular weight \(M\)
The formula relating \(\Delta T\), \(K_b\), weight of solute \(W\), and molecular weight \(M\) is:
\[ \Delta T = \frac{1000 \times K_b \times W}{M \times 83.4} \]
Rearranging to find \(M\):
\[ M = \frac{1000 \times K_b \times W}{\Delta T} \]
Substituting the known values:
- \(K_b = 2.77 \, \text{K kg/mol}\)
- \(W = 1 \, \text{g}\) (assuming 1 g of solute)
- \(\Delta T = 0.175 \, K\)

Calculating \(M\):
\[ M = \frac{1000 \times 2.77 \times 1}{0.175} \approx 158.29 \, \text{g/mol} \]

**Hint**: Make sure to keep track of your units when substituting into the formula.

### Final Answer
The molecular weight of the solute is approximately \(158.29 \, \text{g/mol}\).