Answer: C. Find moles of C and H in the combustion products.
\(n(\text{CO}_2) = 1.65 / 44 = 0.0375\) mol → \(n(\text{C}) = 0.0375\) mol.
\(n(\text{H}_2\text{O}) = 0.90 / 18 = 0.0500\) mol → \(n(\text{H}) = 2 \times 0.0500 = 0.100\) mol.
Ratio C : H = 0.0375 : 0.100 = 3 : 8.
For a monohydric alcohol \(\text{C}_n\text{H}_{2n+1}\text{OH}\): with 3 carbons and 8 hydrogens, n = 3 → \(\text{C}_3\text{H}_7\text{OH}\). Check: 3 C, 7 + 1 = 8 H. ✓ This matches propan-1-ol or propan-2-ol — option C.
Options A (\(\text{CH}_3\text{OH}\)) and B (\(\text{C}_2\text{H}_5\text{OH}\)) have the wrong C:H ratio. Option D (\(\text{C}_3\text{H}_6\text{OH}\)) is not a real alcohol formula — the H count doesn't satisfy \(\text{C}_n\text{H}_{2n+1}\text{OH}\).
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