Answer the following: of a compound is dissolved in of benzene and the freezing point of solution…
Answer the following:
(i)[1.0]
$0\cdot680\text{ g}$ of a compound is dissolved in $15\cdot0\text{ g}$ of benzene and the freezing point of solution is lowered by $1\cdot44^\circ\text{C}$.
Calculate the experimental molecular mass of the compound.
($K_f$ for benzene $= 5\cdot12\text{ K kg mol}^{-1}$)
(ii)[1.0]
If the theoretical molecular mass of the compound referred to above is $80\cdot5\text{ g mol}^{-1}$, suggest whether it is undergoing association or dissociation.
Answer
Answer (i)
AIWritten by AI (gemini) - it can contain mistakes.
Formula:
$M_2 = \frac{K_f \times w_2 \times 1000}{\Delta T_f \times w_1}$
Substitution:
$M_2 = \frac{5\cdot12\text{ K kg mol}^{-1} \times 0\cdot680\text{ g} \times 1000\text{ g kg}^{-1}}{1\cdot44\text{ K} \times 15\cdot0\text{ g}}$
$= \frac{3481\cdot6}{21\cdot6} = 161\cdot19\text{ g mol}^{-1}$
Final answer: 161.19 g mol^-1
Final answer: 161.19 g mol^-1
Answer (ii)
AIWritten by AI (gemini) - it can contain mistakes.
Since the experimental molecular mass ($161\cdot19\text{ g mol}^{-1}$) is greater than the theoretical molecular mass ($80\cdot5\text{ g mol}^{-1}$), the van 't Hoff factor $i = \frac{M_{\text{normal}}}{M_{\text{observed}}} = \frac{80\cdot5}{161\cdot19} \approx 0\cdot5 < 1$. Therefore, the compound is undergoing association (specifically dimerization).
From ISC 2026 Chemistry Paper 1, question 3.