The octane number of a fuel whose performance is the same as that of a mixture of 55 g of 2,2,4- trimethly pentane and 45 g of n-heptane is ?

 The octane number of a fuel whose performance is the same as that of a mixture of 55 g of 2,2,4- trimethly pentane and 45 g of n-heptane is ?

• A. 45
• B. 55
• C. 80
• D. 100

Answer: B. 55

🔹 Step 1: Recall definition of octane number

• Octane number is the percentage by volume of iso-octane (2,2,4-trimethylpentane) in a mixture with n-heptane that has the same knocking performance as the fuel being tested.

• By definition:

$$\text{Octane number} = \% \, \text{iso-octane in the mixture}$$

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🔹 Step 2: Given mixture

• 55 g of iso-octane (2,2,4-trimethylpentane)

• 45 g of n-heptane

Molar masses:

• Iso-octane (C₈H₁₈) = $(12 \times 8) + (1 \times 18) = 114 \, g/mol$

• n-Heptane (C₇H₁₆) = $(12 \times 7) + (1 \times 16) = 100 \, g/mol$

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🔹 Step 3: Convert mass to moles

$$n_{\text{iso-octane}} = \frac{55}{114} \approx 0.482$$ $$n_{\text{heptane}} = \frac{45}{100} = 0.45$$

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🔹 Step 4: Find mole fraction (≈ volume fraction)

$$x_{\text{iso-octane}} = \frac{0.482}{0.482 + 0.45} = \frac{0.482}{0.932} \approx 0.517$$

So iso-octane = 51.7% (≈ 52%).

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🔹 Step 5: Approximation to given options

Closest to 55%.

Or

To determine the octane number of the fuel, we need to consider the composition of the reference mixture.

The reference mixture is made up of:

- 55 g of 2,2,4-trimethylpentane (iso-octane), which has an octane number of 100

- 45 g of n-heptane, which has an octane number of 0

The total mass of the mixture is 55 g + 45 g = 100 g.

Since the performance of the fuel is the same as that of this mixture, we can calculate the octane number of the fuel as follows:

Octane number = (mass of iso-octane / total mass) x 100

Octane number = (55 g / 100 g) x 100

Octane number = 55

Therefore, the correct answer is:

B. 55

Jamb 1990 past question 48

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