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Colligative properties
Appears in
Concepts tested here
- Depression of freezing point
- van't Hoff factor
- van't Hoff factor and freezing point depression
- van't Hoff factor from freezing point depression
All Questions
2014 AIPMT 1 question
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Of the following 0.10 m aqueous solutions, which one will exhibit the largest freezing point depression?Depression in freezing point ∝ van't Hoff factor (i).
For $Al_2(SO_4)_3$, i = 5, the largest among the given solutes.
2011 AIPMT-MAINS 1 question
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A 0.1 molal aqueous solution of a weak acid is 30% ionized. If $K_f$ for water is $1.86^\circ C/m$, the freezing point of the solution will beA weak monobasic acid ionises as $HA \rightleftharpoons H^+ + A^-$, giving n = 2 particles per molecule.
Degree of ionisation $\alpha = 0.30$.
van't Hoff factor: $i = 1 + (n - 1)\alpha = 1 + (2 - 1)(0.30) = 1.3$
$\Delta T_f = i\times K_f\times m = 1.3\times1.86\times0.1$
$\Delta T_f = 0.2418^\circ C$
Freezing point of the solution = $0 - 0.24 = -0.24^\circ C$
2011 AIPMT-PRE 1 question
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The freezing point depression constant for water is $-1.86^\circ C\ m^{-1}$. If 5.00 g $Na_2SO_4$ is dissolved in 45.0 g $H_2O$, the freezing point is changed by $-3.82^\circ C$. Calculate the van't Hoff factor for $Na_2SO_4$.$\Delta T_f = i\times K_f\times m$
Molar mass of $Na_2SO_4$ = 2(23) + 32 + 4(16) = 142 g $mol^{-1}$
Moles of $Na_2SO_4$ = $\frac{5.00}{142} = 0.0352$
Molality $m = \frac{0.0352}{45.0}\times1000 = 0.7825$ mol $kg^{-1}$
$i = \frac{\Delta T_f}{K_f\times m} = \frac{3.82}{1.86\times0.7825}$
$i = \frac{3.82}{1.455} = 2.63$
2010 AIPMT-PRE 1 question
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A solution of sucrose (molar mass = 342 g $mol^{-1}$) has been prepared by dissolving 68.5 g of sucrose in 1000 g of water. The freezing point of the solution obtained will be ($K_f$ for water = 1.86 K kg $mol^{-1}$)Moles of sucrose = $\frac{68.5}{342} = 0.2003$ mol
Molality: $m = \frac{0.2003\ mol}{1\ kg} = 0.2003$ mol $kg^{-1}$
Sucrose is a non-electrolyte (i = 1): $\Delta T_f = K_f\times m = 1.86\times0.2003$
$\Delta T_f = 0.372$ K
Freezing point of solution = $0 - 0.372$
$= -0.372^\circ C$