2.2 g of an alcohol (A) when treated with `CH_3` -Mgl liberates 560 mL of `CH_4` at STP. Alcohol (A) on dehydration followed by ozonolysis gives ketone (B) along with (C). Oxime of ketone (B) contains 19.17% N. (A) on oxidation gives ketone (D) having same number of carbon atom.
Molecular mass of (A) is

Options

A. 74
B. 88
C. 60
D. 102

Explanation

To determine the molecular mass of alcohol (A), we will follow these steps:

### Step 1: Calculate the number of moles of methane (CH₄) produced.
Given that 560 mL of CH₄ is liberated at STP, we can use the molar volume of a gas at STP, which is 22.4 L (or 22400 mL) to find the number of moles.

\[ \text{Number of moles of } CH_4 = \frac{\text{Volume of } CH_4}{\text{Molar volume at STP}} = \frac{560 \, \text{mL}}{22400 \, \text{mL/mol}} = 0.025 \, \text{mol} \]

### Step 2: Relate the moles of alcohol (A) to the moles of methane (CH₄).
From the reaction of alcohol (A) with CH₃MgI, we know that 1 mole of alcohol produces 1 mole of methane. Therefore, the moles of alcohol (A) will also be 0.025 mol.

\[ \text{Moles of A} = 0.025 \, \text{mol} \]

### Step 3: Calculate the molecular mass of alcohol (A).
We are given that the mass of alcohol (A) is 2.2 g. To find the molecular mass, we can use the formula:

\[ \text{Molecular mass of A} = \frac{\text{mass of A}}{\text{moles of A}} = \frac{2.2 \, \text{g}}{0.025 \, \text{mol}} = 88 \, \text{g/mol} \]

### Conclusion:
The molecular mass of alcohol (A) is **88 g/mol**.

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