What weight of solte (mol.wt.60) is required to dissolve in 180g of water to reduce the vapour pressure to `4//5th` of pure water?

Explanation

To solve the problem of determining the weight of solute required to reduce the vapor pressure of water to \( \frac{4}{5} \) of its pure value, we can use the concept of relative lowering of vapor pressure. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Problem
We need to find the weight of a solute (molecular weight = 60 g/mol) that must be dissolved in 180 g of water to achieve a vapor pressure that is \( \frac{4}{5} \) of the vapor pressure of pure water.

### Step 2: Define Variables
- Let \( P_0 \) be the vapor pressure of pure water.
- The vapor pressure of the solution \( P_s \) is given as \( P_s = \frac{4}{5} P_0 \).
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