The vapour pressure of a solution of 5g of non-electrolyte in 100g of water at a particular temperature is `2985 N m^(-2)`. The vapour pressure of pure water at that temperature is `3000 N m^(-2)`. The molecular weight of the solute is

Options

A. 180
B. 90
C. 270
D. 200

Explanation

To find the molecular weight of the solute, we can use the concept of relative lowering of vapor pressure according to Raoult's law. Here’s a step-by-step solution:

### Step 1: Understand the given data
- Vapor pressure of pure water (P₀) = 3000 N/m²
- Vapor pressure of the solution (P₁) = 2985 N/m²
- Mass of the solute (non-electrolyte) = 5 g
- Mass of the solvent (water) = 100 g
- Molecular weight of water (Mₕ₂ₒ) = 18 g/mol

### Step 2: Calculate the lowering of vapor pressure
The lowering of vapor pressure (ΔP) can be calculated as:
\[ ΔP = P₀ - P₁ = 3000 \, \text{N/m}² - 2985 \, \text{N/m}² = 15 \, \text{N/m}² \]

### Step 3: Calculate the relative lowering of vapor pressure
The relative lowering of vapor pressure (R) is given by:
\[ R = \frac{ΔP}{P₀} = \frac{15 \, \text{N/m}²}{3000 \, \text{N/m}²} = \frac{1}{200} \]

### Step 4: Relate relative lowering to mole fraction of solute
According to Raoult's law, the relative lowering of vapor pressure is equal to the mole fraction of the solute (Xₛ):
\[ R = Xₛ = \frac{nₛ}{nₛ + nₕ₂ₒ} \]
Where:
- \(nₛ\) = number of moles of solute
- \(nₕ₂ₒ\) = number of moles of solvent (water)

### Step 5: Calculate the number of moles of solvent
The number of moles of water (nₕ₂ₒ) can be calculated using its mass and molecular weight:
\[ nₕ₂ₒ = \frac{100 \, \text{g}}{18 \, \text{g/mol}} = \frac{100}{18} \approx 5.56 \, \text{mol} \]

### Step 6: Set up the equation for mole fraction
Substituting the values into the equation for mole fraction:
\[ \frac{1}{200} = \frac{nₛ}{nₛ + 5.56} \]

### Step 7: Express moles of solute in terms of its molecular weight
Let the molecular weight of the solute be M. The number of moles of solute (nₛ) is given by:
\[ nₛ = \frac{5 \, \text{g}}{M} \]

### Step 8: Substitute and solve for M
Substituting \(nₛ\) into the mole fraction equation:
\[ \frac{1}{200} = \frac{\frac{5}{M}}{\frac{5}{M} + 5.56} \]

Cross-multiplying gives:
\[ 5.56 \cdot 1 = \frac{5}{M} \cdot 200 \Rightarrow 5.56 = \frac{1000}{M} \]

Rearranging gives:
\[ M = \frac{1000}{5.56} \approx 180.18 \, \text{g/mol} \]

### Final Answer
The molecular weight of the solute is approximately **180.18 g/mol**.

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