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Q 1 :
What happens to freezing point of benzene when small quantity of naphthalene is added to benzene? [2024]

Increases
Remains unchanged
First decreases and then increases
Decreases

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2

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4

(D)

When a small quantity of naphthalene is added to benzene its freezing point decreases.

Q 2 :
The osmotic pressure of a dilute solution is $7 \times 10^{5}$ Pa at 273 K. Osmotic pressure of the same solution at 283 K is _______ $\times 10^{4} {Nm}^{- 2}$ . [2024]

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(73)

$Osmotic pressure(π) (for a non - electrolyte solution) is related to concentration (C) and temperature (T) as,$

$π = C R T, where R is universal gas constant$

$π_{1} = C R T_{1}, π_{2} = C R T_{2}$

$\frac{π_{2}}{π_{1}} = \frac{T_{2}}{T_{1}}$

$\frac{π_{2}}{7 \times 10^{5} Pa} = \frac{283 K}{273 K}$

$π_{2} = 7 \times 10^{5} \times \frac{283}{273} Pa$

$= 7.256 \times 10^{5} {Nm}^{- 2}$

$= 72.56 \times 10^{4} {Nm}^{- 2}$

Q 3 :
Mass of ethylene glycol (antifreeze) to be added to 18.6 kg of water to protect the freezing point at -24°C is _____ kg.

(Molar mass in g ${mol}^{- 1}$ for ethylene glycol 62, $K_{f}$ of water = 1.86 K kg ${mol}^{- 1}$ ) [2024]

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(15)

$Freezing point of pure water T_{f}^{o} = 0 ° C$

$Freezing point of ethylene glycol solution T_{f} = - 24 ° C$

$Depression in freezing point, Δ T_{f} = T_{f}^{o} - T_{f} = 0 ° C - (- 24 ° C) = 24 ° C$

$Value of difference of two temperatures is same in ° C and K, thus$

$Δ T_{f} = 24 ° C or 24 K$

$Δ T_{f} = i \times m \times k_{f}$

$Assuming glycol to be undissociated in water, i = 1$

$24 = 1 \times m \times 1.86$

$m = \frac{24}{1.86}$

$\frac{W_{glycol}}{M_{glycol} \times W_{water} (in kg)} = \frac{24}{1.86}$

$\frac{W_{glycol}}{62 \times 18.6} = \frac{24}{1.86}$

$W_{glycol} = 14880 g = 14.88 kg$

Q 4 :
2.5 g of a non-volatile, non-electrolyte is dissolved in 100 g of water at 25°C. The solution showed a boiling point elevation by 2°C. Assuming the solute concentration in negligible with respect to the solvent concentration, the vapour pressure of the resulting aqueous solution is ____ mm of Hg (nearest integer).

[Given: Molal boiling point elevation constant of water $(K_{b}) = 0.52 K . kg {mol}^{- 1}$ ,1 atm pressure = 760 mm of Hg, Molar mass of water = 18 g ${mol}^{- 1}$ ] [2024]

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(711)

$Δ T_{b} = m \times k_{b}$

$2 = m \times 0.52$

$m = \frac{2}{0.52} m$

$\frac{x_{solute} \times 1000}{x_{solvent} \times M_{solvent}} = \frac{2}{0.52}$

$\frac{x_{solute} \times 1000}{x_{solvent} \times 18} = \frac{2}{0.52}$

$\frac{x_{solute}}{x_{solvent}} = \frac{2}{0.52} \times \frac{18}{1000}$

$\frac{x_{solvent}}{x_{solute}} = \frac{130}{9}$

$x_{solvent} = \frac{130}{9 + 130} = \frac{130}{139}$

By Raoult's law:

$Δ T_{b} = m \times k_{b}$

$P_{solution} = P_{solvent}^{\circ} x_{solvent}$

$= 760 \times \frac{130}{139} mmHg$

$= 710.8 mmHg \approx 711 mmHg$

Note: $P_{solvent}^{\circ}$ is not given in question, question is solved by taking it as 760 mmHg which is not correct as 760 mmHg is vapor pressure of water at $100 ° C$ but temperature in question is $25 ° C$ .

Q 5 :
2.7 Kg of each of water and acetic acid are mixed. The freezing point of the solution will be $- x$ °C. Consider the acetic acid does not dimerise in water, nor dissociates in water. $x$ = _______ (nearest integer)

[Given: Molar mass of water = 18 g ${mol}^{- 1}$ , acetic acid = 60 g ${mol}^{- 1}$

$K_{f} H_{2} O : 1.86 K k g m o l^{- 1}$

$K_{f} acetic acid : 3.90 K kg {mol}^{- 1}$

$Freezing point: H_{2} O = 273 K, acetic acid = 290 K$ ] [2024]

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(31)

$Moles of water = \frac{given mass}{molar mass} = \frac{2700}{18} = 150 mol$

$Moles of acetic acid = \frac{given mass}{molar mass} = \frac{2700}{60} = 45 mol$

As water is in excess, it is solvent.

$Molality (m) = \frac{n_{acetic acid}}{W_{water} (in kg)} = \frac{45}{2.7} m = 16.667 m$

$Δ T_{f} = m k_{f} (solvent) = m k_{f} (water)$

$= 16.667 \times 1.86 ° C = 31 ° C$

$T_{f} (water) - T_{f} (solution) = 31 ° C$

$0 ° C - T_{f} (solution) = 31 ° C$

$T_{f} (solution) = - 31 ° C$

Q 6 :
An artificial cell is made by encapsulating 0.2 M glucose solution within a semipermeable membrane. The osmotic pressure developed when the artificial cell is placed within a 0.05 M solution of NaCl at 300 K is ______ $\times 10^{- 1}$ bar. (Nearest Integer)

[Given: R = 0.083 L bar ${mol}^{- 1}$ $K^{- 1}$ ]

Assume complete dissociation of NaCl [2024]

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(25)

Osmotic pressure developed = Difference of osmotic pressure due to glucose solution difference of osmotic pressure due to NaCl solution.

$= C_{1} R T - i C_{2} R T$

$= 0.2 R T - 2 \times 0.05 R T$

$= 0.1 R T = 0.1 mol \times 0.083 \frac{bar L}{mol K} \times 300 K = 25 \times 10^{- 1} bar$

Q 7 :
Considering acetic acid dissociates in water, its dissociation constant is $6.25 \times 10^{- 5}$ . If 5 mL of acetic acid is dissolved in 1 litre water, the solution will freeze at $- x \times 10^{- 2}$ °C, provided pure water freezes at 0 °C. $x$ = ________ .(Nearest integer)

Given:    ${(K_{f})}_{water}$ = 1.86 K kg ${mol}^{- 1}$ .

density of acetic acid is 1.2 g ${mol}^{- 1}$ .

molar mass of water = 18 g ${mol}^{- 1}$ .

molar mass of acetic acid = 60 g ${mol}^{- 1}$ .

density of water = 1 g ${cm}^{- 3}$

  Acetic acid dissociates as

   ${CH}_{3} COOH ⇌ {CH}_{3} {COO}^{⊝} + H^{\oplus}$ [2024]

0%

(19)

$W_{{CH}_{3} COOH} = density \times volume = 1.2 \frac{g}{mL} \times 5 mL = 6 g$

$W_{H_{2} O} = density \times volume = 1 \frac{g}{mL} \times 1000 mL = 1000 g$

$m = \frac{W_{{CH}_{3} COOH} \times 1000}{M_{{CH}_{3} COOH} \times W_{water} (in g)} = \frac{6 \times 1000}{60 \times 1000 (in g)} m = 0.1 m$

${CH}_{3} COOH ⇌ {CH}_{3} {COO}^{-} + H^{+}$

$K_{a} = \frac{C α^{2}}{1 - α} \approx C α^{2}$

$α = \sqrt{\frac{K_{a}}{C}} = \sqrt{\frac{6.25 \times 10^{- 5}}{0.1}} = 0.025$

$i = 1 + (n - 1) α = 1 + (2 - 1) \times 0.025 = 1.025$

$Δ T_{f} = i m k_{f} = 1.025 \times 0.1 \times {1.86}^{\circ} C = {0.19}^{\circ} C$

$T_{f} ° - T_{f} = {0.19}^{\circ} C$

$0 - T_{f} = {0.19}^{\circ} C$

$T_{f} = - {0.19}^{\circ} C = 19 \times {10^{- 2}}^{\circ} C$

Q 8 :
Consider the dissociation of the weak acid HX as given below:

$HX(aq) ⇌ H^{+} (aq) + X^{-} (aq), K a = 1.2 \times 10^{- 5}$

[ $K_{a}$ : dissociation constant]

The osmotic pressure of 0.03 M aqueous solution of HX at 300 K is ______ $\times 10^{- 2}$ bar (nearest integer).

[Given: R = 0.083 L bar ${mol}^{- 1}$ $K^{- 1}$ ] [2024]

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(76)

$HX(aq) ⇌ H^{+} (aq) + X^{-} (aq)$

$(t = 0)$ $C$

$(c o n c .)$

$(t = \infty)$ $C (1 - α)$ $C α$ $C α$

$(c o n c .)$

$K_{a} = \frac{C α^{2}}{1 - α} \approx C α^{2}$

$1.2 \times 10^{- 5} = 0.03 α^{2}$

$α = \sqrt{\frac{1.2 \times 10^{- 5}}{0.03}} = 0.02$

$i = 1 + (n - 1) α = 1 + (2 - 1) \times 0.02 = 1.02$

$π = i C R T = 1.02 \times 0.03 \frac{mol}{L} \times 0.083 \frac{bar L}{mol K} \times 300 K = 76.2 \times 10^{- 2} bar$

Topic Question Set

Q 1 :
What happens to freezing point of benzene when small quantity of naphthalene is added to benzene? [2024]

Q 2 :
The osmotic pressure of a dilute solution is $7 \times 10^{5}$ Pa at 273 K. Osmotic pressure of the same solution at 283 K is _______ $\times 10^{4} {Nm}^{- 2}$ . [2024]

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Q 3 :
Mass of ethylene glycol (antifreeze) to be added to 18.6 kg of water to protect the freezing point at -24°C is _____ kg.

(Molar mass in g ${mol}^{- 1}$ for ethylene glycol 62, $K_{f}$ of water = 1.86 K kg ${mol}^{- 1}$ ) [2024]

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Topic Question Set

Q 1 : What happens to freezing point of benzene when small quantity of naphthalene is added to benzene? [2024]

Q 2 : The osmotic pressure of a dilute solution is 7×105 Pa at 273 K. Osmotic pressure of the same solution at 283 K is _______ ×104 Nm-2. [2024]

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Q 3 : Mass of ethylene glycol (antifreeze) to be added to 18.6 kg of water to protect the freezing point at -24°C is _____ kg. (Molar mass in g mol-1 for ethylene glycol 62, Kf of water = 1.86 K kg mol-1) [2024]

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Q 8 : Consider the dissociation of the weak acid HX as given below: HX(aq)⇌H+(aq)+X-(aq), Ka=1.2×10-5 [Ka: dissociation constant] The osmotic pressure of 0.03 M aqueous solution of HX at 300 K is ______ ×10-2 bar (nearest integer). [Given: R = 0.083 L bar mol-1 K-1] [2024]

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Q 1 :
What happens to freezing point of benzene when small quantity of naphthalene is added to benzene? [2024]

Q 2 :
The osmotic pressure of a dilute solution is $7 \times 10^{5}$ Pa at 273 K. Osmotic pressure of the same solution at 283 K is _______ $\times 10^{4} {Nm}^{- 2}$ . [2024]

Q 3 :
Mass of ethylene glycol (antifreeze) to be added to 18.6 kg of water to protect the freezing point at -24°C is _____ kg.

(Molar mass in g ${mol}^{- 1}$ for ethylene glycol 62, $K_{f}$ of water = 1.86 K kg ${mol}^{- 1}$ ) [2024]