If 30g of a solute of molecular weight 154 is dissolved in 250g of benzene, what will be the boiling point of the resulting solution under atmospheric pressure . The molal boiling point elevation constant for benzene is `2.61^(@)C.m^(-1)` and the boiling point of pure benzene is `80.1^(@)C`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the number of moles of solute
To find the number of moles of the solute, we use the formula:
\[ \text{Moles of solute} = \frac{\text{mass of solute (g)}}{\text{molecular weight of solute (g/mol)}} \]
Given:
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