Chemistry 5 Marks Questions

1. It shows positive deviation.

It is due to weaker interaction between acetone and ethanol than ethanol-ethanol interactions.

2. Given: WB = 10g WS = 100g, WA = 90g MB = 180g/mol & d = 1.2g/m L

$\text{M}=\frac{\text{Wt}\text{ %}\times\text{density}\times\text{10}}{\text{Mol.wt}}$

$\text{M}=\frac{\text{10}\times{1.2}\times{10}}{{180}}=0.66\text{M }\text{ or }\text{ 0.66 mol/L}$

$\text{m}=\frac{\text{W}_{B}\times{1000}}{\text{M}_{B}\times{W}_{A}\text{(in g)}}$

$\text{m}=\frac{\text{10}\times{1000}}{\text{180}\times\text{90}}$

= 0.61m or 0.61mol/kg.

1. $\triangle\text{T}_{\text{f}}=\text{K}_{\text{f}}\text{m}$

$\text{Here}, \ \text{m}=\frac{\text{w}_2\times1000}{\text{M}_2\times\text{M}_1}$

$273.15-269.15=\frac{\text{K}_{\text{f}}\times10\times1000}{342\times90}$

$\text{K}_{\text{f}}=12.3\text{ }\text{K}\text{ }\text{kg}/\text{mol}$

$\triangle\text{T}_{\text{f}}=\text{K}_{\text{f}}\text{m}$

$=\frac{12.3\times10\times1000}{180\times90}$

$=7.6\text{ }\text{K}$

$\text{T}_{\text{f}}=273.15-7.6=265.55\text{ }\text{K}$

1. Number of moles of solute dissolved in per kilo gram of the solvent.
2. Abnormal molar mass: If the molar mass calculated by using any of the colligative properties to be different than theoretically expected molar mass.

1. $\frac{(\text{P}_{\text{A}}^0-\text{P}_{\text{A}})}{\text{P}_\text{A}^0}=\frac{(\text{w}_{\text{B}}\times\text{M}_{\text{A}})}{(\text{M}_{\text{B}}\times\text{w}_{\text{A}})}$

$\frac{23.8-\text{P}_{\text{A}}}{23.8}=30\times\frac{18}{60}\times846$

$23.8-\text{P}_{\text{A}}=23.8\times[30\times\frac{18}{60}\times846]$

$23.8-\text{P}_{\text{A}}=0.2532$

$\text{P}_{\text{A}}=23.55\text{ }mm\text{ }\text{H}g$

2. Ideal solution	Non ideal solution
   (i) It obeys Raoult’s law over the entire range of concentration. (ii) Δ𝑚𝑖𝑥 𝐻 = 0 (iii) Δ𝑚𝑖𝑥 𝑉 = 0	(i) Does not obey Raoult’s law over the entirerange of concentration. (ii) Δ𝑚𝑖𝑥 𝐻 is not equal to 0. (iii) Δ𝑚𝑖𝑥 𝑉 is not equal to 0.

1. Molarity is defined as number of moles of solute dissolved in one litre of solution.
2. It is equal to elevation in boiling point of 1 molal solution.

2. For isotonic solutions: $\pi$ urea = $\pi$ glucose

$\frac{\text{W}_\text{urea}}{\text{M}_\text{urea}\times\text{V}_{s}}=\frac{\text{W}_{Glucose}}{\text{M}_{Glucose}\times\text{V}_{s}}$ (As volume of solution is same)

$ \frac{\text{W}_{urea}}{\text{M}_{urea}}=\frac{\text{W}_{Glucose}}{\text{M}_{Glucose}}$ or $ \frac{\text{15g}}{\text{60g mol}^{-1}}=\frac{\text{W}_{Glucose}}{\text{180g mol}^{-1}}$

$\text{W}_{Glucose}=\frac{\text{15g}\times\text{180g mol}^{-1}}{\text{60g mol}^{-1}}=\text{45g}$

1. Partial vapour pressure of a liquid component is directly proportional to its mole fraction in its solution.

The partial pressure of the volatile component or gas is directly proportional to its mole fraction in solution. Only the proportionality constant K_H differs from P^o_A. Thus, Raoult’s law becomes a special case of Henry’s law in which K_H becomes equal to P^o_A.

 $\Delta\text{T}_{f}=\text{K}_{f}\frac{\text{W}_{B}\times{1000}}{\text{M}_{B}\times\text{W}_{A}\text{(in grams)}}$

$\text{M}_{B}=\text{K}_{f}\frac{\text{W}_{B}\times{1000}}{\Delta\text{T}_{f}\times\text{W}_{A}}$

$\text{M}_{B}=\frac{\text{5.12}\times{1}\times{1000}}{\text{0.40}\times\text{50}}$

= 256g mol^-1

1. Ideal Solution: Those solutions which follows Raoult’s law under all conditions of temperature and pressure.
2. Azeotrope: A liquid mixture which distills at constant temperature without under going any change in composition is called Azeotrope.
3. Osmotic Pressure: The minimum excess pressure that has to be applied on the solution side to prevent the entry of the solvent in to the solution through the semi - permeable membrane is called osmotic pressure.

2. Given Molecular mass of Glucose = 180, % by wt = 10

$\text{M}=\frac{\text{1000}\times\text{wt%}}{\text{(100-wt%)}\times\text{mol. wt. of solute}}$ OR $\text{M}=\frac{\text{w}\times\text{1000}}{\text{M}\times\text{W}}$

$\text{M}=\frac{\text{1000}\times\text{10}}{\text{(100-10)}\times\text{180}}$

$\text{M}=\frac{\text{10000}}{\text{90}\times\text{180}}$

m = 0∙617m.

1. The Ratio of number ofmoles of one component to the total number of moles of solution./ormathematical expression.
2. The Solution which follows Raoult’s law over the entire range of concentrations.

2.  WB = 15 g    WA = 450 g
   $\Delta$T_f = 0.34 ^oC    K_f = 1.86 Kkg/mol    MB = ?

$\text{M}_{B}=\frac{1000\times\text{K}_{f}\times\text{W}_{B}}{\Delta\text{T}_{f}\times\text{W}_{A}}$

$\text{M}_{B}=\frac{1000\times\text{1.86 k mol}^{-1}\times\text{15g}}{0.34\text{K}\times\text{450g}}$

= 182.35 g/mol.

1. The Partial pressure of the gas above the liquid is directly proportional to the mole fraction of the gas dissolved in the liquid.
2. Boiling PointElevation Constant. It is equal to elevation in boiling point of 1 molal solution, i.e., 1 mole of solute is dissolved in 1 kg of solvent.

2. W_B = ?    W_A = 500 g    $\Delta$T_b = 100.42 ^oC – 100 ^oC = 0.42 ^oC or 0.42K
   K_b = 0.512 K kg/mol      M_B = 92g/mol

$\Delta\text{T}_{b}=\text{K}_{b}\frac{\text{W}_{B}\times1000}{\text{M}_{B}\times\text{W}_{A{\text{(in gram)}}}}$

$\text{W}_{B}=\frac{\Delta\text{T}_{b}\times\text{W}_{\text{A(in grams)}}}{\text{1000}\times\text{K}_{b}}$

$=\frac{\text{042K}\times\text{92 g mol}^{-1}\times\text{500kg}}{\text{1000}\times\text{0.512K kg mol}^{-1}}$

WB = 37.73 g

1. Mole fraction is the ratio of number of moles of one component to the total number of moles in a mixture.
2. Van't Hoff factor is expressed as:

$\text{i}=\frac{\text{norml molar mass (m}_{1})}{\text{abnormal molar mass (m'}_{1})\text{(due to dissociation or association )}}$

2. $\pi$ = CRT

$\text{M}_{2}=\frac{\text{w2RT}}{\pi\text{ V}}$

$\text{M}_{2}=\frac{100\times10^{-3}\text{g}\times0.0821\text{ L atm mol}^{-1}\text{K}^{-1}\times\text{298 K}\times760\times1000}{\text{13.3 atm}\times\text{10L}}$

M₂ = 13980g mol^–1 or 1.4 x 10⁴g mol^–1.

1. Properties, whose values depend only on the concentration (number) of solute particles in solution and not on the nature of the solute, are called colligative properties.
2. Molality (m) is the Number of moles of solute dissolved in one kg of the solvent.

2. Given p_(gas) = 0.78 atm = 0.78 x 760 mm = 592.8 mm Hg

K_H = 8.42 x 10^–7 M/mm Hg.

Xgas = ?

X gas = K_H$\times$P

X gas = 8.42 x 10^–7 x 592.8

= 4991.376 x 10^–7 = 0.4991 x 10^–3

$\text{X}_{N2}=\frac{\text{n}_{N_2}}{\text{n}_{N_2}+\text{n H}_{2}\text{O}}$

0.4991x10^–3 = $\frac{\text{n}_{N_2}}{\text{n H}_{2}\text{O}}$

0.4991x10^–3 = $\frac{\text{n}_{N_2}}{\text{55.55}}$

$\text{n}_{\text{N}_{2}}$ = 0.4991 X 10^–3 X 55.55 = 2.772 x 10^–3 M.

1. $ \Delta\text{T}_\text{f}=\frac{\text{K}_\text{f}{W}_\text{b}\times{1000}}{\text{M}_\text{b}\times{W}_\text{a}}$

0.383 =( 3.83×2.56/M×100) ×1000
M=256
S × x = 256
32× x= 256
x=8.

1. Shrinks.
2. Swells.

1. $ \Delta\text{T}_\text{f}=\text{i}\frac{\text{K}_\text{f}{W}_\text{b}\times{1000}}{\text{M}_\text{b}\times{W}_\text{a}}$

$\Delta$T_f = 3×(1.86×1.9/95×50) ×1000
= 2.23K
T_f -$\Delta$T_f’ = 273.15 - 2.23/273 - 2.23
T_f’ = 270.92 K or 270.77K.

2. 2M glucose; More number of particles/less vapour pressure.
3. Reverse Osmosis.

1. The flow of solvent molecules from solution of low concentration to higher concentration through semipermeable membrane is called osmosis. 

The hydrostatic pressure that has to be applied on the solution to prevent the entry of the solvent into the solution through the semipermeable membrane is called the Osmotic Pressure.

Yes osmotic pressure is a colligative property as it depends upon the number of particles of the solute in a solution.

2. $\text{T}_{b}=\text{iK}_{b}\text{ m}$

$\text{T}_{b}-\text{T}_{b}^{0}=2\times 0.512\text{K kg mol}^{-1}\times\frac{15 g}{\text{58.44 gmol}^{-1}}\times\frac{1000}{250kg}$

T_b-373 K = 1.05 K
T_b = 374.05K or 101.05^oC.

1. Molality (m) is the number of moles of the solute per kilogram (kg) of the solvent whereas Molarity is the number of moles of solute present in one litre (or one cubic decimeter) of solution at a particular temperature. 

Molality is independent of temperature whereas Molarity is function. of temperature because volume depends on temperature and the mass does not or Molarity decreases with increase of temperature.

2. $\Delta\text{T}_{f}=7.5^{o}C$

$\Delta\text{T}_{f}=\text{iK}_{f}\text{m}$

$\text{T}_{f}^{0}-\text{T}_{f}=3\times1.86^0\text{C kg mol}^{-1}\times\frac{10.50g}{\text{184 gmol}^{-1}}\times\frac{\text{1000}}{\text{200 kg}}$

0^oC-T_f = 1.59^oC
T_f = –1.59^oC or 271.41 K.

$\therefore\ \frac{\text{n}_\text{B}}{\text{n}_\text{A}+\text{n}_\text{B}}=1-0.85=0.15\ \dots(\text{ii})$

Dividing (ii) by (i), we get

$\frac{\text{n}_\text{B}}{\text{n}_\text{A}}=\frac{0.15}{0.85}$

or $\frac{\text{n}_\text{B}}{1000/18}=\frac{0.15}{0.85}$

or $\text{n}_\text{B}=\frac{0.15}{0.85}\times\frac{1000}{18}=9.8\text{ mol}$

Hence, molality = 9.8m

Explanation:

1. Raoult's law: Mathematical respresentation of Raoult's law $\text{p}=\text{x}_1\text{p}^0_1+\text{x}_2\text{p}^0_2$

2. Henry's law: $\text{p}=\text{K}_\text{H}\cdot\text{x}$

3. Elevation of boiling point: Mathematical representation, $\Delta\text{T}_\text{b}=\text{K}_\text{b}\text{m}$

4. Depression in freezing point: Mathematical representation, $\Delta\text{T}_\text{f}=\text{K}_\text{f}\text{m}$

5. Osmotic pressure: Mathematical representation, $\pi=\text{CRT}$

1. w/w (mass percentage) $=\frac{\text{Mass of component in the solution}}{\text{Total mass of the solution}}\times100$

2. V/V (volume percentage) $=\frac{\text{Volume of the component}}{\text{Total volume of the solution}}\times100$

3. w/V (mass by volume percentage) $=\frac{\text{Mass of solute}}{100\text{ml of solution}}$

4. ppm. (parts per million) $=\frac{\text{Number of parts of component}}{\text{Total number of parts of all components of the solution}}\times10^{6}$

5. x (mole fraction) $=\frac{\text{Number of moles of component}}{\text{Total number of moles all the components}}$

6. M (Molarity) $=\text{M}=\frac{\text{Moles of solute}}{\text{Volume of solution in litre}}$

7. m (Molality) $=\frac{\text{Moles of solute}}{\text{Mass of solvent in kg}}$

8. Mass percentage, ppm, mole fraction and molality are independent of temperature since mass does not change with temperature.

Now, consider the following equilibrium for the acid:

$\text{i}=\frac{\text{Total number of mole so f particles after association}}{\text{Number of mole so f particles be fore association}}$

$\text{i}=\frac{1-\frac{\alpha}{2}}{1}=1-\frac{\alpha}{2}\ \dots(\text{ii})$

Now, according to Henry's law:

$\text{x}_{\text{O}_2}=\frac{\text{P}_{\text{O}_2}}{\text{K}_\text{H}}$

$=\frac{1520\text{ mm Hg}}{3.30\times10^7\text{ mm Hg}(\text{Given K}_\text{H}=3.30\times10^7\text{mm Hg})}$

$\text{x}_{\text{N}_2}=\frac{\text{P}_{\text{N}_2}}{\text{K}_\text{H}}$

$=\frac{6004\text{ mm Hg}}{6.51\times10^7\text{ mm Hg}}$

Hence, the mole fractions of oxygen and nitrogen in water are 4.61 × 10^-5 and 9.22 × 10^-5 respectively.

Therefore, No. of moles present in 10 g of CH₃CH₂CHClCOOH $=\frac{10\text{ g}}{122.5\text{ g mol}^{-1}}$

It is given that 10 g of CH₃CH₂CHClCOOH is added to 250 g of water.

$\text{Initial conc. At equilibrium}=\text{CH}_3\text{CH}_2\text{CHClCOOH}\leftrightarrow\text{CH}_3\text{CH}_2\text{CHClCOOH}^-+\text{H}^+\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{C mol L}^{-1}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{C}(1-\alpha)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{C}\alpha\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{C}\alpha$

$\text{K}_\text{a}=\frac{\text{C}\alpha.\text{C}\alpha}{\text{C}(1-\alpha)}$

$=\frac{\text{C}\alpha^2}{1-\alpha}$

$\text{K}_\text{a}=\frac{\text{C}\alpha^2}{1}$

$\text{Now,}\text{ K}_\text{a}=\text{C}\alpha^2$

$\alpha=\sqrt{\frac{\text{K}_\text{a}}{\text{C}}}$

$=\sqrt{\frac{1.4\times10^{-3}}{0.3264}}(\because\ \text{K}_\text{a}=1.4\times10^{-3})$

$\text{Initial moles at equilibrium}=\text{CH}_3\text{CH}_2\text{CH}\text{ClCOOH}\leftrightarrow\text{CH}_3\text{CH}_2\text{CH}\text{ClCOOH}^-+\text{H}^+\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1-\alpha\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \alpha\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \alpha$

$\text{Total moles of equilibrium = 1− α + α + α}$

$=1+\alpha$

$\text{Therefore,}\text{ i}=\frac{1+\alpha}{1}$

$=1+\alpha$

= 1.0655 × 1.86 K kg mol^-1 × 0.3264 mol kg^-1

$=0.575\times5\times0.4=1.15\text{K}$

$\text{T}_{\text{f}}=\text{T}^0_\text{f}-\Delta\text{T}_\text{f}$

$=278.4\text{ K}-1.15\text{ K}=277.25\text{ K}$

Van't Hoff Factor $(\text{i})=\frac{\text{Observed freezing point}}{\text{Calculated freezing point}}=\frac{0.0205\text{ K}}{0.0197\text{ K}}$

i = 1.041 ...(i)

1. Association: Compounds like benzoic acid or ethanoic acid dimerise in benzene due to hydrogen bonding as a result of which the number of particles in the solution decreases. Since colligative properties depend upon number of particles, such solutes show lower colligative property.

2. Dissociation: Electrolytes like NaCl, KCl, etc. dissociate into ions which result in increased number of particles, hence higher value of colligative property.

To account for association or dissociation van’t Hoff introduced a factor T known as van’t Hoff factor. It is defined as Expected molar mass Abnormal molar mass Observed colligative property Calculated colligative property. Total number of moles of particles after association/ dissociation Total number of moles of particles before association/ dissociation.

$\text{i}=\frac{\text{Expected molar mass}}{\text{Abnormal molar mass}}$

$\text{i}=\frac{\text{Observed colligative property}}{\text{Calculated colligative property}}$

$\text{i}=\frac{\text{Total number of moles of particles}\\\text{after association/ dissociation}}{\text{Total number of moles of particles}\\\text{before association/ dissociation}}$

$\text{We know that:}\ \text{ M}_2=\frac{\text{K}_\text{f}\times\text{w}_2\times1000}{\Delta\text{T}_\text{f}\times\text{w}_1}$

$=\frac{1.86\text{ K kg mol}^{-1}\times19.5\text{ g}\times1000\text{ g Kg}^{-1}}{500\text{ g}\times1\text{ K}}$