At `334K` the vapour pressure of benzene`(C_(6)H_(6))`is`0.526`atm and that of toluene`(C_(7)H_(8))`is`0.188`atm. In a solution containing`0.500`mole of benzene and`0.500`mole of toluene , what is the vapour pressure of toluene for above solution at`334K` ?

Options

A. 0.188 atm
B. 0.10 atm
C. 0.357 atm
D. 0.094 atm

Answer

To find the vapor pressure of toluene in a solution containing equal moles of benzene and toluene at 334K, we will use Raoult's Law. Here’s a step-by-step solution:

### Step 1: Identify the given data
- Vapor pressure of pure benzene (P0B) = 0.526 atm
- Vapor pressure of pure toluene (P0T) = 0.188 atm
- Moles of benzene (nB) = 0.500 moles
- Moles of toluene (nT) = 0.500 moles

### Step 2: Calculate the total moles in the solution
Total moles (n_total) = nB + nT = 0.500 + 0.500 = 1.000 moles

### Step 3: Calculate the mole fractions of benzene and toluene
- Mole fraction of benzene (XB) = nB / n_total = 0.500 / 1.000 = 0.500
- Mole fraction of toluene (XT) = nT / n_total = 0.500 / 1.000 = 0.500

### Step 4: Apply Raoult's Law to find the partial vapor pressure of toluene
According to Raoult's Law, the partial vapor pressure of toluene (PT) in the solution can be calculated as:
\[ PT = XT \times P0T \]

Substituting the values:
\[ PT = 0.500 \times 0.188 \text{ atm} \]
\[ PT = 0.094 \text{ atm} \]

### Step 5: Conclusion
The vapor pressure of toluene in the solution at 334K is **0.094 atm**.

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