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Solve the following.
An aqueous solution of a non-volatile solute freezes at , while pure water freezes at . Determine…
(i)[2.5]
An aqueous solution of a non-volatile solute freezes at $272\cdot4\text{ K}$, while pure water freezes at $273\cdot0\text{ K}$. Determine the following:
(Given $K_f = 1\cdot86\text{ K kg mol}^{-1}$, $K_b = 0\cdot512\text{ K kg mol}^{-1}$ and vapour pressure of water at $298\text{ K} = 23\cdot756\text{ mm of Hg}$)
(1) The molality of solution
(2) Boiling point of solution
(3) The lowering of vapour pressure of water at $298\text{ K}$
(ii)[2.5]
A solution containing $1\cdot23\text{ g}$ of calcium nitrate in $10\text{ g}$ of water, boils at $100\cdot975^\circ\text{C}$ at $760\text{ mm of Hg}$. Calculate the van’t Hoff factor for the salt at this concentration.
($K_b$ for water $= 0\cdot52\text{ K kg mol}^{-1}$, mol. wt. of calcium nitrate $= 164\text{ g mol}^{-1}$)
Answer
Answer (i)
AIFormulas used: Depression of freezing point: dTf = i*Kf*m with Kf = 1.86 K kg mol^-1, dTf = 0.6 K, so m = 0.3226 mol kg^-1 Elevation of boiling point: dTb = i*Kb*m with m = 0.3226 mol kg^-1, Kb = 0.512 K kg mol^-1, so dTb = 0.1652 K
$\Delta T_f = 273\cdot0 - 272\cdot4 = 0\cdot6\text{ K}$
(1) $m = \dfrac{\Delta T_f}{K_f} = \dfrac{0\cdot6}{1\cdot86} = 0\cdot323\text{ mol kg}^{-1}$
(2) $\Delta T_b = K_b m = 0\cdot512\times0\cdot323 = 0\cdot165\text{ K}$, so boiling point $= 373\cdot15 + 0\cdot165 = 373\cdot315\text{ K}$ ($100\cdot165^\circ\text{C}$)
(3) $\dfrac{p^\circ - p}{p^\circ} = x_2 = \dfrac{0\cdot323}{55\cdot5 + 0\cdot323} = 5\cdot77\times10^{-3}$, so $p^\circ - p = 23\cdot756\times5\cdot77\times10^{-3} = 0\cdot137\text{ mm of Hg}$
Final answer: 0.323 mol/kg; 373.315 K; 0.137 mm Hg mol kg^-1; K; mm of Hg
Answer (ii)
AIFormula used: Elevation of boiling point: dTb = i*Kb*m with Kb = 0.52 K kg mol^-1, M2 = 164 g mol^-1, w1 = 10 g, w2 = 1.23 g, dTb = 0.975 K, so i = 2.5
$\Delta T_b = 100\cdot975 - 100 = 0\cdot975\text{ K}$
$\Delta T_b = i\,K_b\,\dfrac{w_2\times1000}{M_2\times w_1}$, so $0\cdot975 = i\times0\cdot52\times\dfrac{1\cdot23\times1000}{164\times10}$
$0\cdot975 = i\times0\cdot52\times0\cdot75$, giving $i = 2\cdot5$
Final answer: 2.5
From ISC 2020 Chemistry Paper 1, question 16(b).
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