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Answer the following.
of a compound is dissolved in of benzene and the freezing point of solution is lowered by …
$0\cdot680\text{ g}$ 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 $= 5\cdot12\text{ K kg mol}^{-1}$)
Answer
Answer
AIFormula:
$M_2 = \frac{K_f \times w_2 \times 1000}{\Delta T_f \times w_1}$
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 $\text{g mol}^{-1}$
From ISC 2026 Chemistry Paper 1, question 3(i).
Check your working with the Molar mass calculator: molar mass of any formula, hydrates too, with the working and percentage composition.
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