- A${C_5}{H_{10}}{O_5}$
- B${C_3}{H_4}{O_3}$
- C${C_{12}}{H_{22}}{O_{11}}$
- ✓${C_6}{H_{12}}{O_6}$
$\therefore $ $1$ mole carbohydrate has hydrogen $ = \frac{1}{{0.0833}} = 12\,g$
Empirical Formula $(C{H_2}O)$ has hydrogen $ = 2\,g$
Hence $n = \frac{{12}}{2} = 6$
Hence molecular formula of carbohydrate $ = {(C{H_2}O)_6} = {C_6}{H_{12}}{O_6}$
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$A$. A liquid evaporates to vapour.
$B$. Temperature of a crystalline solid lowered from $130 \mathrm{~K}$ to $0 \mathrm{~K}$.
$C$. $2 \mathrm{NaHCO}_{3(\mathrm{~s})} \rightarrow \mathrm{Na}_2 \mathrm{CO}_{3(\mathrm{~s})}+\mathrm{CO}_{2(\mathrm{~g})}+\mathrm{H}_2 \mathrm{O}_{(\mathrm{g})}$
$D$. $\mathrm{Cl}_{2(g)} \rightarrow 2 \mathrm{Cl}_{(g)}$
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