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Solve the following.
The elevation in boiling point when of acetic acid is dissolved in of benzene is . Calculate the…
(i)[2.5]
The elevation in boiling point when $0\cdot30\text{ g}$ of acetic acid is dissolved in $100\text{ g}$ of benzene is $0\cdot0633^\circ\text{C}$. Calculate the molecular weight of acetic acid from this data. What conclusion can you draw about the molecular state of the solute in the solution?
(Given $K_b$ for benzene $= 2\cdot53\text{ K kg mol}^{-1}$, at. wt. of $\text{C} = 12$, $\text{H} = 1$, $\text{O} = 16$)
(ii)[2.5]
Determine the osmotic pressure of a solution prepared by dissolving $0\cdot025\text{ g}$ of $\text{K}_2\text{SO}_4$ in $2\text{ litres}$ of water at $25^\circ\text{C}$, assuming that $\text{K}_2\text{SO}_4$ is completely dissociated.
($R = 0\cdot0821\text{ Lit-atm K}^{-1}\text{mol}^{-1}$, mol. wt. of $\text{K}_2\text{SO}_4 = 174\text{ g mol}^{-1}$)
Answer
Answer (i)
AIFormula used: Elevation of boiling point: dTb = i*Kb*m with Kb = 2.53 K kg mol^-1, w1 = 100 g, w2 = 0.30 g, dTb = 0.0633 K, so M2 = 119.9 g mol^-1
$M_2 = \dfrac{K_b\times w_2\times1000}{\Delta T_b\times w_1} = \dfrac{2\cdot53\times0\cdot30\times1000}{0\cdot0633\times100} = 119\cdot9\text{ g mol}^{-1}$
True molecular weight of $\mathrm{CH_3COOH}$ = $2\times12 + 4\times1 + 2\times16 = 60\text{ g mol}^{-1}$.
Conclusion: the observed value is about double the normal value, so acetic acid associates (forms dimers) in benzene.
Final answer: 119.9 g mol^-1
Answer (ii)
AIFormula used: Osmotic pressure: pi = i*C*R*T with R = 0.0821 L atm K^-1 mol^-1, T = 298 K, V = 2 L, i = 3, M2 = 174 g mol^-1, w2 = 0.025 g, so pi = 0.00527 atm
$\mathrm{K_2SO_4} \rightarrow 2\mathrm{K^+} + \mathrm{SO_4^{2-}}$, so $i = 3$
$\pi = iCRT = i\,\dfrac{w_2}{M_2 V}RT = 3\times\dfrac{0\cdot025}{174\times2}\times0\cdot0821\times298 = 5\cdot27\times10^{-3}\text{ atm}$
Final answer: 0.00527 atm
From ISC 2020 Chemistry Paper 1, question 16(a).
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