JEE Chemistry Solutions Notes Set 04

Concentration Terms

 

Term	Definition	Formula
Mass percent (w/w)	Grams of solute in 100 g of solution	\( \dfrac{w_{solute}}{w_{solution}}\times100 \)
Mole fraction	Moles of a component / total moles	\( x_A=\dfrac{n_A}{n_A+n_B} \), sum of all x = 1
Molarity (M)	Moles of solute per litre of solution; changes with temperature	\( M=\dfrac{n}{V_{L}} \)
Molality (m)	Moles of solute per kg of solvent; independent of temperature	\( m=\dfrac{n}{w_{solvent}(kg)} \)
ppm / ppb	Parts of solute per \( 10^6 \) / \( 10^9 \) parts of solution by mass	\( \dfrac{w_{solute}}{w_{solution}}\times10^6 \)

 

Conversions: \( M=\dfrac{10\,d\,(\%w/w)}{M_{solute}} \) with d in g/mL, and \( m=\dfrac{1000\,M}{1000\,d-M\,M_{solute}} \).

 

Example 1. 18 M \( \ce{H2SO4} \) has density 1.8 g/cm\(^3\). Mass of 1 L solution = 1800 g; acid = 18 x 98 = 1764 g, water = 36 g. So molality = 18/0.036 = 500 mol/kg, mass percent = 1764/1800 = 98%, mole fraction of acid = 18/(18+2) = 0.9 and of water 0.1.

 

Vapour Pressure and Raoult's Law

 

A liquid in a closed vessel reaches equilibrium with its vapour; the pressure of that vapour is the vapour pressure. It depends only on the nature of the liquid and the temperature, not on the amount of liquid, the shape of the vessel or the volume above the liquid. It rises with temperature (Clausius-Clapeyron equation) and falls with stronger intermolecular forces. A liquid boils when its vapour pressure equals the external pressure; the normal boiling point is the temperature at which the vapour pressure is 1 atm.

 

Raoult's law: the partial vapour pressure of a volatile component in a solution equals its mole fraction multiplied by its pure vapour pressure, \( p_A=x_Ap_A^0 \). For a binary solution of two volatile liquids, \( P_T=p_A^0x_A+p_B^0x_B=p_B^0+(p_A^0-p_B^0)x_A \), a straight line against \( x_A \). The vapour composition follows Dalton's law, \( y_A=p_A/P_T \), and the vapour is richer in the more volatile component.

 

For a non-volatile solute: the relative lowering of vapour pressure equals the mole fraction of the solute.

 

\[ \dfrac{p^0-p_s}{p^0}=x_{solute}=\dfrac{n}{n+N},\qquad \dfrac{p^0-p_s}{p_s}=\dfrac{n}{N} \]

 

For solutes that ionise or associate replace n by \( i\,n \).

 

Example 2. Vapour pressure of pure water is 23.76 mm and of a solution of 20 g solute in 100 g water is 22.41 mm. Then \( n/N=(23.76-22.41)/22.41=0.0602 \), N = 5.556, n = 0.3347 mol, M = 20/0.3347 = 59.8 g/mol.

 

Gases in Liquids and Immiscible Liquids

 

Henry's law: at constant temperature the mass or mole fraction of a gas dissolved in a liquid is proportional to its partial pressure, \( p=K_Hx \). A larger \( K_H \) means lower solubility. Solubility of gases decreases as temperature rises because dissolution is exothermic. In terms of logs, \( \log m=\log K+\log p \) is a straight line of slope 1.

 

Immiscible liquids each exert their own full vapour pressure, so \( P_T=p_A^0+p_B^0 \) and the mixture boils below the boiling point of either liquid (the basis of steam distillation). In the vapour, \( \dfrac{n_A}{n_B}=\dfrac{p_A^0}{p_B^0} \) and \( \dfrac{w_A}{w_B}=\dfrac{p_A^0M_A}{p_B^0M_B} \).

 

Ideal and Non-Ideal Solutions

 

Feature	Ideal	Positive deviation	Negative deviation
Raoult's law	Obeyed at all concentrations	Vapour pressure higher than ideal	Vapour pressure lower than ideal
A-B compared with A-A and B-B	Equal	A-B weaker	A-B stronger
\( \Delta H_{mix} \)	0	Positive	Negative
\( \Delta V_{mix} \)	0	Positive	Negative
Azeotrope	None	Minimum boiling	Maximum boiling
Examples	Benzene-toluene, n-hexane-n-heptane, chlorobenzene-bromobenzene	Ethanol-water, acetone-carbon disulphide, ethanol-cyclohexane	Chloroform-acetone, water-HNO3, water-HCl

 

An azeotrope is a mixture that boils at constant temperature with the same composition in liquid and vapour, so it cannot be separated by fractional distillation. For every ideal solution \( \Delta G_{mix}<0 \) and \( \Delta S_{mix}>0 \).

 

Colligative Properties

 

Properties that depend only on the number of solute particles in a given amount of solvent, not on their nature, are colligative: relative lowering of vapour pressure, elevation of boiling point, depression of freezing point and osmotic pressure. They hold for dilute solutions of non-volatile solutes.

 

Property	Expression	Constant
Relative lowering of vapour pressure	\( \dfrac{p^0-p}{p^0}=x_{solute} \)	-
Elevation of boiling point	\( \Delta T_b=iK_bm \)	\( K_b=\dfrac{RT_b^2M_{solvent}}{1000\,\Delta H_{vap}} \)
Depression of freezing point	\( \Delta T_f=iK_fm \)	\( K_f=\dfrac{RT_f^2M_{solvent}}{1000\,\Delta H_{fus}} \)
Osmotic pressure	\( \pi=iCRT \)	R = 0.0821 L atm mol-1 K-1

 

Water has \( K_b=0.52 \) and \( K_f=1.86 \) K kg/mol. Molar mass of a solute from a measured effect: \( M=\dfrac{1000\,K\,w_{solute}}{\Delta T\;w_{solvent}(g)} \).

 

van't Hoff Factor

 

The van't Hoff factor is the ratio of the observed colligative effect to the calculated effect for no dissociation or association, \( i=\dfrac{\text{observed number of particles}}{\text{initial number}}=\dfrac{\text{normal molar mass}}{\text{observed molar mass}} \).

 

Process	Degree alpha	i
Dissociation into n ions, \( A\to n \) particles	\( \alpha \)	\( 1+(n-1)\alpha \)
Association, \( nA\to A_n \)	\( \alpha \)	\( 1-\alpha+\dfrac{\alpha}{n} \)

 

For complete dissociation i equals the number of ions: NaCl 2, \( \ce{CaCl2} \) 3, \( \ce{K3[Fe(CN)6]} \) 4, \( \ce{K4[Fe(CN)6]} \) 5. For a complex, only ionisable (outer-sphere) ions count, so \( \ce{[Pt(NH3)4Cl2]Cl2} \) gives 3 ions.

 

Example 3. A 0.01 molal ammonia solution freezes at -0.02 degrees C: \( i=0.02/(1.86\times0.01)=1.075 \), so \( \alpha=0.075 \).

 

Osmosis

 

Osmosis is the spontaneous flow of solvent through a semipermeable membrane from the dilute solution (or pure solvent) to the concentrated solution. The pressure that has to be applied to the solution side to stop the flow is the osmotic pressure. Solutions of equal osmotic pressure at the same temperature are isotonic; a solution with higher pressure is hypertonic and with lower pressure hypotonic. Reverse osmosis (applying more than the osmotic pressure) is used for desalination. Osmotic pressure is the colligative property most suited to polymers and proteins because the effect is large even for dilute solutions.

 

Example 4. For 3.00 g of \( \ce{Ca(NO3)2} \) in 100 mL at 285 K with \( \pi=11.2 \) atm: \( C=0.183 \) M, \( i=11.2/(0.183\times0.0821\times285)=2.62 \), and \( \alpha=(2.62-1)/2=0.81 \).

 

Example 5. Isotonic solutions: 1.0 M NaCl (i = 2) and 1.0 M \( \ce{K2SO4} \) of unknown ionisation. Since \( i\,C \) must be equal, i = 2 for \( \ce{K2SO4} \), giving \( 1+2\alpha=2 \) and \( \alpha=50\% \).