A solution containing $1.8 \mathrm{~g}$ of a compound (empirical formula $\mathrm{CH}_{2} \mathrm{O}$ ) in…

A solution containing $1.8 \mathrm{~g}$ of a compound (empirical formula $\mathrm{CH}_{2} \mathrm{O}$ ) in $40 \mathrm{~g}$ of water is observed to freeze at $-0.465^{\circ} \mathrm{C}$. The molecular formula of the compound is $\left(\mathrm{K}_{\mathrm{f}}ight.$ of water $\left.=1.86 \mathrm{~kg} \mathrm{~K} \mathrm{~mol}^{-1}ight)$
  1. $\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}_{2}$
  2. $\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}_{3}$
  3. $\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}_{4}$
  4. $\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}$

Solution

$\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{K}_{\mathrm{f}} \times \mathrm{m}$
$\mathrm{M}=\frac{1000 \times \mathrm{K}_{\mathrm{f}} \times \mathrm{w}_{2}(\text { solute })}{\Delta \mathrm{T}_{\mathrm{f}} \times \mathrm{w}_{1}(\text { solvent })}$
$=\frac{1000 \times 1.86 \times 1.8}{0.465 \times 40} \Rightarrow \mathrm{M}=180$
Molecular formula $=(\text { empirical formula })_{n}$ $\mathrm{n}=\frac{\text { Molecular mass }}{\text { Empirical formula mass }}=\frac{180}{30}=6$
Molecular formula $=\left(\mathrm{CH}_{2} \mathrm{O}ight)_{6}=\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}$

Asked in: JEE-TOPICTESTS-CHEMISTRY

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