At `50^(@)C` the vapour pressures of pure water and ethyl alcohol are, respectively `92.5mm` and `219.9mmHg`. If 6g of nonvolatile solute of mol.wt. `120s` is dissolved in `150g` of each of these solvents, what will be the relative vapour pressure lowerings in the two solvents ?

Answer

To solve the problem of relative vapor pressure lowering in two solvents (water and ethyl alcohol) when a non-volatile solute is dissolved, we will follow these steps:

### Step 1: Calculate the moles of the solute
Given:
- Mass of solute = 6 g
- Molecular weight of solute = 120 g/mol

The number of moles of the solute can be calculated using the formula:
\[ \text{Moles of solute} = \frac{\text{mass of solute}}{\text{molecular weight of solute}} = \frac{6 \text{ g}}{120 \text{ g/mol}} = 0.05 \text{ moles} \]

### Step 2: Calculate the moles of ethyl alcohol and water
**For Ethyl Alcohol:**
- Mass of ethyl alcohol = 150 g
- Molecular weight of ethyl alcohol = 46 g/mol

\[ \text{Moles of ethyl alcohol} = \frac{150 \text{ g}}{46 \text{ g/mol}} \approx 3.26 \text{ moles} \]

**For Water:**
- Mass of water = 150 g
- Molecular weight of water = 18 g/mol

\[ \text{Moles of water} = \frac{150 \text{ g}}{18 \text{ g/mol}} \approx 8.33 \text{ moles} \]

### Step 3: Calculate the mole fractions of the solute in both solvents
**For Ethyl Alcohol:**
\[ \text{Mole fraction of solute in ethyl alcohol} = \frac{\text{moles of solute}}{\text{moles of solute} + \text{moles of ethyl alcohol}} = \frac{0.05}{0.05 + 3.26} \approx 0.015 \]

**For Water:**
\[ \text{Mole fraction of solute in water} = \frac{\text{moles of solute}}{\text{moles of solute} + \text{moles of water}} = \frac{0.05}{0.05 + 8.33} \approx 0.006 \]

### Step 4: Calculate the relative vapor pressure lowering for both solvents
The relative lowering of vapor pressure (RLVP) can be expressed as:
\[ \text{RLVP} = \text{Mole fraction of solute} \]

Thus, we have:
- For Ethyl Alcohol: RLVP = 0.015
- For Water: RLVP = 0.006

### Step 5: Compare the relative vapor pressure lowerings
From the calculations:
- Relative vapor pressure lowering in ethyl alcohol is greater than that in water.

### Final Answer:
The relative vapor pressure lowering in ethyl alcohol is approximately 0.015, while in water it is approximately 0.006. Therefore, the relative vapor pressure lowering is greater in ethyl alcohol.

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