Question

A sample of 4.12 mg of unknown alcohol is added to \(CH_3MgBr\). 1.56 mL of methane at STP was liberated. The alcohol is :

• A. \(C_2H_5OH\)
• B. \(CH_3OH\)
• C. \(C_3H_7OH\)
• D. \(C_4H_9OH\)

Hint:

Always convert mg → g before calculating molar mass.

Answer 1

The Correct Option is C

Concept: \[ ROH + CH_3MgBr \rightarrow RH + CH_4 \] 1 mol alcohol gives 1 mol methane.

Step 1: Moles of methane.
\[ n = \frac{1.56}{22400} = 6.96 \times 10^{-5} \approx 7 \times 10^{-5} \]

Step 2: Moles of alcohol.
\[ n_{alcohol} = n_{CH_4} = 7 \times 10^{-5} \]

Step 3: Molar mass.
\[ M = \frac{\text{mass}}{\text{moles}} = \frac{4.12 \times 10^{-3}}{7 \times 10^{-5}} \] \[ M \approx 58.9 \approx 60 \]

Step 4: Identify alcohol.
Molar mass $\approx 60 \Rightarrow C_3H_7OH$