A mixture of two immiscible liquids nitrobenzene and water boiling point at `99^(@)` C has a partial vapour pressure of water 733 mm and that of nitro benzene 27mm. Calculate the ratio of the weights of water to nitro benzene in the distillate.

Answer

To solve the problem of finding the ratio of the weights of water to nitrobenzene in the distillate, we can use Raoult's law and the relationship between partial vapor pressures and mole fractions. Here’s a step-by-step solution:

### Step 1: Understand the given data
We have the following information:
- Partial vapor pressure of water (Pw) = 733 mm
- Partial vapor pressure of nitrobenzene (Pn) = 27 mm

### Step 2: Apply Raoult's Law
According to Raoult's law, the partial vapor pressure of a component in a mixture is proportional to its mole fraction in the liquid phase. Therefore, we can write:

\[ \frac{P_w}{P_n} = \frac{x_w}{x_n} \]

Where:
- \(x_w\) = mole fraction of water
- \(x_n\) = mole fraction of nitrobenzene

### Step 3: Express mole fractions
Since the sum of mole fractions in a binary mixture is equal to 1, we can express \(x_n\) in terms of \(x_w\):

\[ x_n = 1 - x_w \]

### Step 4: Substitute into the equation
Now substituting \(x_n\) into the equation from Step 2:

\[ \frac{P_w}{P_n} = \frac{x_w}{1 - x_w} \]

### Step 5: Plug in the values
Substituting the values of \(P_w\) and \(P_n\):

\[ \frac{733}{27} = \frac{x_w}{1 - x_w} \]

### Step 6: Cross-multiply and solve for \(x_w\)
Cross-multiplying gives:

\[ 733(1 - x_w) = 27x_w \]

Expanding this:

\[ 733 - 733x_w = 27x_w \]

Combining like terms:

\[ 733 = 733x_w + 27x_w \]
\[ 733 = 760x_w \]

Now, solving for \(x_w\):

\[ x_w = \frac{733}{760} \]

### Step 7: Calculate \(x_n\)
Now we can find \(x_n\):

\[ x_n = 1 - x_w = 1 - \frac{733}{760} = \frac{760 - 733}{760} = \frac{27}{760} \]

### Step 8: Relate mole fractions to weights
The number of moles can be expressed in terms of weight and molar mass:

\[ N_w = \frac{W_w}{M_w} \quad \text{and} \quad N_n = \frac{W_n}{M_n} \]

Where:
- \(W_w\) = weight of water
- \(W_n\) = weight of nitrobenzene
- \(M_w\) = molar mass of water (approximately 18 g/mol)
- \(M_n\) = molar mass of nitrobenzene (approximately 123 g/mol)

### Step 9: Set up the ratio of weights
Using the mole fractions:

\[ \frac{N_w}{N_n} = \frac{x_w}{x_n} \]

Substituting the expressions for \(N_w\) and \(N_n\):

\[ \frac{\frac{W_w}{M_w}}{\frac{W_n}{M_n}} = \frac{x_w}{x_n} \]

This simplifies to:

\[ \frac{W_w}{W_n} = \frac{x_w \cdot M_n}{x_n \cdot M_w} \]

### Step 10: Substitute the values
Substituting \(x_w\) and \(x_n\):

\[ \frac{W_w}{W_n} = \frac{\left(\frac{733}{760}\right) \cdot 123}{\left(\frac{27}{760}\right) \cdot 18} \]

The \(760\) cancels out:

\[ \frac{W_w}{W_n} = \frac{733 \cdot 123}{27 \cdot 18} \]

### Step 11: Calculate the ratio
Calculating the right-hand side:

\[ \frac{W_w}{W_n} = \frac{733 \cdot 123}{27 \cdot 18} = \frac{90159}{486} \approx 185.5 \]

### Final Step: Present the ratio
Thus, the ratio of the weights of water to nitrobenzene in the distillate is approximately:

\[ \frac{W_w}{W_n} \approx 185.5 \]