jee main chemistry important questions | Solutions jee main questions FREE

Q 11 :

The vapour pressure vs. temperature curve for a solution–solvent system is shown below:

The boiling point of the solvent is ______ °C. [2023]

(82)

The temperature at which vapour pressure of solvent becomes 1 atm is called normal B.P. of solvent. So, the boiling point of the solvent is 82°C.

Q 12 :

If the boiling points of two solvents X and Y (having same molecular weights) are in the ratio 2 : 1 and their enthalpy of vaporization are in the ratio 1 : 2, then the boiling point elevation constant of X is m times the boiling point elevation constant of Y. The value of m is ________ (Nearest Integer) [2023]

(8)

$Δ T_{b} = K_{b} m$

$K_{b} = \frac{R T_{b}^{2} \times M_{solvent}}{1000 \times Δ H_{vap}}$

$\frac{{(Δ T_{b})}_{_{X}}}{{(Δ T_{b})}_{Y}} = {[\frac{{(T_{b})}_{X}}{{(T_{b})}_{Y}}]}^{2} \times \frac{M_{X}}{M_{Y}} \times \frac{Δ H_{Y}}{Δ H_{X}} = {(\frac{2}{1})}^{2} (\frac{1}{1}) \times \frac{2}{1} = 8$

Q 13 :

An aqueous solution of volume 300 ${cm}^{3}$ contains 0.63 g of protein. The osmotic pressure of the solution at 300 K is 1.29 mbar. The molar mass of the protein is ________ g ${mol}^{- 1}$ . Given: R = 0.083 L bar $K^{- 1}$ ${mol}^{- 1}$ [2023]

(40535)

$π = \frac{W / M}{V} R T$

or $M = \frac{W \times R \times T}{π \times V}$ $= \frac{0.63 \times 0.083 \times 300}{1.29 \times 10^{- 3} \times 300 \times 10^{- 3}}$

or $M = 40535 g/mol$

Q 14 :

Solutions of 12 g of non-electrolyte (A) prepared by dissolving it in 1000 mL of water exerts the same osmotic pressure as that of 0.05 M glucose solution at the same temperature. The empirical formula of A is ${CH}_{2} O$ . The molecular mass of A is ______ g. (Nearest integer) [2023]

(240)

$For isotonic solutions$

$π_{1} = π_{2}$

$C_{1} = C_{2} (For non-electrolyte)$

$(\frac{12}{M_{A} \times 1}) = 0.05$

$M_{A} = 240$

Q 15 :

A solution is prepared by dissolving 0.3 g of a non-volatile non-electrolyte solute A of molar mass 60 g ${mol}^{- 1}$ and 0.9 g of a non-volatile non-electrolyte solute B of molar mass 180 g ${mol}^{- 1}$ in 100 mL $H_{2} O$ at 27°C. Osmotic pressure of the solution will be:

[Given: R = 0.082 L atm $K^{- 1}$ ${mol}^{- 1}$ ] [2026]

1.23 atm
1.47 atm
2.46 atm
0.82 atm

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(3)

$Mass of solute 'A' = 0.3 g$

$Moles of solute 'A' = \frac{0.3 g}{60 {g mol}^{- 1}} = \frac{1}{200} mol$

$Mass of solute 'B' = 0.9 g$

$Moles of solute 'B' = \frac{0.9 g}{180 {g mol}^{- 1}} = \frac{1}{200} mol$

$Total molarity of all solutes = \frac{2 / 200}{100} \times 1000 = \frac{1}{10} M$

$∴$ $π = \frac{1}{10} \times 0.082 \times 300$

$π = 2.46 atm$

Q 16 :

Elements P and Q form two types of non-volatile, non-ionizable compounds PQ and ${PQ}_{2} .$ When 1 g of PQ is dissolved in 50 g of solvent ‘A’, ${ΔT}_{b}$ was 1.176 K while when 1 g of ${PQ}_{2}$ is dissolved in 50 g of solvent 'A', ${ΔT}_{b}$ was 0.689 K. $(K_{b} of ‘A’ = 5 {K kg mol}^{- 1})$ . The molar masses of elements P and Q $(in g mol⁻¹)$ respectively, are: [2026]

65, 145
70, 110
60, 25
25, 60

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(4)

${(Δ T_{b})}_{P Q} = K_{b} m$

$1.176 = 5 \times \frac{1}{M_{1}} \times \frac{1000}{50}$

$M_{1} = 85.03$

${(Δ T_{b})}_{P Q_{2}} = 5 \times \frac{1}{M_{2}} \times \frac{1000}{50} = 0.689$

$M_{2} = 145.13$

$Let molar mass of P & Q are M_{P} and M_{Q} respectively$

$M_{P} + M_{Q} = 85.03$

$M_{P} + 2 M_{Q} = 145.13$

$M_{P} = 24.93 \approx 25$

$M_{Q} = 60.1 \approx 60$

Q 17 :

Consider the following aqueous solutions. [2026]

I. 2.2 g Glucose in 125 mL of solution.
II. 1.9 g Calcium chloride in 250 mL of solution.
III. 9.0 g Urea in 500 mL of solution.
IV. 20.5 g Aluminium sulphate in 750 mL of solution.

The correct increasing order of boiling point of these solutions will be:
[Given : Molar mass in g ${mol}^{- 1}$ : H = 1, C = 12, N = 14, O = 16, Cl = 35.5, Ca = 40, Al = 27 and S = 32]

III < I < II < IV
I < II < III < IV
II < III < I < IV
II < III < IV < I

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(2)

$Δ T_{b} = i \cdot k_{b} \cdot m$

For dilute solution $(M = m)$

	Molarity	$i \times m$
(I)	$M_{glucose} = \frac{2.2}{180} \times \frac{1000}{125} = 0.098$	$0.098 \times 1$
(II)	$M_{{CaCl}_{2}} = \frac{1.9}{111} \times \frac{1000}{250} = 0.068$	$0.068 \times 3$
(III)	$M_{urea} = \frac{9}{60} \times \frac{1000}{500} = 0.3$	$0.3 \times 1$
(IV)	${M_{{Al}_{2} ({SO}_{4})}}_{3} = \frac{20.5}{342} \times \frac{1000}{750} = 0.08$	$0.08 \times 5$

$Order of Δ T_{b} = {Al}_{2} {({SO}_{4})}_{3} > Urea > {CaCl}_{2} > Glucose$

$So order of BP = {Al}_{2} {({SO}_{4})}_{3} > Urea > {CaCl}_{2} > Glucose$

$So Answer will be I < II < III < IV$

Q 18 :

A substance ‘X’ (1.5 g) dissolved in 150 g of a solvent ‘Y’ (molar mass = 300 g ${mol}^{- 1}$ ) led to an elevation of the boiling point by 0.5 K. The relative lowering in the vapour pressure of the solvent ‘Y’ is ______ $\times 10^{- 2}$ . (nearest integer)

[Given : $K_{b}$ of the solvent = 5.0 K kg ${mol}^{- 1}$ ]

Assume the solution to be dilute and no association or dissociation of X takes place in solution. [2026]

(3)

$Δ T_{b} = i \times K_{b} \times m$

$0.5 = i \times m \times 5$

$i \times m = \frac{0.5}{5} = 0.1$

$i \times a = \frac{15}{1000}$

(where a = moles of solute)

Now,

$\frac{P ° - P_{s}}{P °} = i X_{solute} = i \times \frac{a}{a + \frac{150}{300}}$

$= i \times \frac{a}{\frac{1}{2}} = \frac{15 / 1000}{1 / 2} = \frac{30}{1000} = 3 \times 10^{- 2} = 3$

Q 19 :

The osmotic pressure of a living cell is 12 atm at 300 K. The strength of sodium chloride solution that is isotonic with the living cell at this temperature is ______ g $L^{- 1}$ . (Nearest integer)

Given : R = 0.08 L atm $K^{- 1} {mol}^{- 1}$

Assume complete dissociation of NaCl

(Given : Molar mass of Na and Cl are 23 and 35.5 g ${mol}^{- 1}$ respectively.) [2026]

(15)

$π = i C R T$

$12 = 2 \times C \times 0.08 \times 300$

$12 = 2 \times C \times 24$

$C = \frac{1}{4} mole/L$

$then strength of NaCl solution$

$= \frac{1}{4} \times 58.5 g/L$

$= 14.625 g/L$

$\approx 15 g/L$

Q 20 :

A sample of sea water was found to contain $9.5 %$ , w/w of $M g C l_{2}$ and $5.85 %$ , w/w of NaCl (Assuming complete ionization of salts) What will be the boiling point of sample?

( $K_{b}$ for water $= 0.52$ ,K-kg/mol, At. Wt of Mg = 24, Na = 23, Cl = 35.5)

$102.6 ° C$
$103.07 ° C$
$101.04 ° C$
$100.52 ° C$

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(2)

$In 100 gm solution$

$n_{{MgCl}_{2}} = 0.1$ $n_{NaCl} = 0.1$

$After dissociation total moles = 0.1 \times 3 + 0.1 \times 2 = 0.5$

$W_{water} = 100 - 9.5 - 5.85 = 84.65 gm$

$∴$ $Δ T_{b} = 0.52 \times 0.5 \times \frac{1000}{84.65} = 3.07 ° C$

$∴$ $T_{b} = 103.07 ° C$

Q 21 :

Molar depression constant for a solvent is $8.0 K kg {mol}^{- 1}$ . The depression in freezing point of the solvent for $0.03 mol {kg}^{- 1}$ solution of $K_{2} S O_{4}$ is ______ ?

(Assume complete dissociation of the electrolyte)

0.72 K
0.24 K
0.36 K
0.12 K

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(1)

$K_{2} S O_{4} \to 2 K^{+} + S O_{4}^{2 -} i = 3$

Q 22 :

The freezing point of equimolal aqueous solutions will be highest for:

$N a C l$
$C a {(N O_{3})}_{2}$
$A l C l_{3}$
$C_{6} H_{12} O_{6}$

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(4)

Depression in freezing point, $Δ T_{f} = i K_{f} m$ . The value of van’t Hoff factor (i) is minimum for the glucose, which is a non-electrolyte. Hence, aqueous solution of glucose has highest freezing point.

Q 23 :

Solution A is prepared by dissolving 1 g of a protein (molar mass $= 50000 {g mol}^{- 1}$ ) in 0.5 L of water at 300 K. Its osmotic pressure is $x$ bar. Solution B is made by dissolving 2 g of same protein in 1 L of water at 300 K. Osmotic pressure of solution B is $y$ bar. Entire solution of A is mixed with entire solution of B at same temperature. The osmotic pressure of resultant solution is $z$ bar. $x$ , $y$ and $z$ respectively are:

( $R = 0.083 {L bar mol}^{- 1} K^{- 1}$ ) [2026]

$9.96 \times 10^{- 4}, 9.96 \times 10^{- 4}, 9.96 \times 10^{- 4}$
$9.96 \times 10^{- 4}, 9.96 \times 10^{- 4}, 19.92 \times 10^{- 4}$
$4.98 \times 10^{- 4}, 4.98 \times 10^{- 4}, 9.96 \times 10^{- 4}$
$4.98 \times 10^{- 4}, 4.98 \times 10^{- 4}, 4.98 \times 10^{- 4}$

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(1)

$Osmotic pressure π = CRT$

Solution A:

$x = \frac{1}{50000 \times 0.5} \times 0.083 \times 300 bar = 9.96 \times 10^{- 4} bar$

Solution B:

$y = \frac{2}{50000 \times 1} \times 0.083 \times 300 bar = 9.96 \times 10^{- 4} bar$

Solution A + Solution B:

$z = \frac{(1 + 2)}{50000 \times (0.5 + 1)} \times 0.083 \times 300 bar = 9.96 \times 10^{- 4} bar$

Q 24 :

At $27 ° C$ , 0.1 M, 1 L $K_{4} [Fe ({CN)}_{6}]$ aqueous solution and 0.1 M, 1 L ${FeCl}_{3}$ aqueous solution are placed in a container separated by a semi permeable membrane $AB$ . Assume complete dissociation of both the solutes. Which of the following statement is correct? [2026]

[IMAGE 202]

Blue color is formed on both sides.
Ionic solutes in aqueous solution can pass through semi-permeable membrane.
Solution on side ' $y$ ' is hypotonic.
To cause the reverse flow of solvent during osmosis, external pressure (any value) should be applied to side ' $x$ '.

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(3)

(1) (2)

$4 {FeCl}_{3} + 3 K_{4} [{Fe(CN)}_{6}] \to \underset{(Prussian blue)}{{Fe}_{4} {[{Fe(CN)}_{6}]}_{3}} + 12 KCl$

Above reaction does not occur as solute molecules do not pass through semi permeable membrane.

(3) $K_{4} [{Fe(CN)}_{6}] \to 4 K^{+} + {[{Fe(CN)}_{6}]}^{4 -} (i = 5)$ , assuming 100% ionization

$π_{K_{4} [{Fe(CN)}_{6}]} = i C R T = 5 \times 0.1 R T = 0.5 R T$

${FeCl}_{3} \to {Fe}^{3 +} + 3 {Cl}^{-} (i = 4), assuming 100 % ionization$

$π_{{FeCl}_{3}} = i C R T = 4 \times 0.1 R T = 0.4 R T$

Solution with lesser osmotic pressure ( $Y$ ) is hypotonic.

(4) For reverse osmosis to occur, pressure to be applied on X side should be more than difference in osmotic pressure of the two compartments.

Q 25 :

A non-volatile, non-electrolyte solid solute when dissolved in 40 g of a solvent, the vapour pressure of the solvent decreased from 760 mm Hg to 750 mm Hg. If the same solution boils at 320 K, then the number of moles of the solvent present in the solution is ______ . (Nearest integer)

[Given: boiling point of the pure solvent = 319.5 K, $K_{b}$ of the solvent $= 0.3 {K kg mol}^{- 1}$ ] [2026]

(5)

$\frac{P_{A} ° - P_{A}}{P_{A} °} = x_{B} (mole fraction of solute)$

$\frac{760 - 750}{760} = x_{B}$

$x_{B} = \frac{1}{76}$

$x_{A} (mole fraction of solvent)$

$= 1 - x_{B} = 1 - \frac{1}{76} = \frac{75}{76}$

$Molality (m) = \frac{x_{B} \times 1000}{x_{A} M_{A}} = Δ T_{b} = m K_{b}$

$= \frac{(1 / 76) \times 1000}{(75 / 76) \times M_{A}} = \frac{1 \times 1000}{75 \times M_{A}}$

$Δ T_{b} = m K_{b}$

$320 - 319.5 = \frac{1 \times 1000}{75 \times M_{A}} \times 0.3$

$M_{A} = 8 {g mol}^{- 1}$

Number of moles of solvent

$= \frac{Given mass}{Molar mass} = (\frac{40}{8}) mol = 5 mol.$

Q 26 :

20 g hemoglobin in a 1 L aqueous solution (A) at 300 K is separated from pure water by semi permeable membrane. At equilibrium the height of solution in a tube dipped in a solution (A) is found to be 80.0 mm higher than the tube dipped in water.

The molar mass of hemoglobin is _______ ${kg mol}^{- 1}$ . (Nearest integer)

(Given: $g = 10 {ms}^{- 2}$ , $R = 8.3 {kPa dm}^{3} K^{- 1} {mol}^{- 1}$ , density of solution = $1000 {kg m}^{- 3}$ ) [2026]

(62)

Osmotic pressure $π = h ρ g$

$= 80 \times 10^{- 3} m \times 1000 {kg m}^{- 3} \times 10 {ms}^{- 2}$

$= 800 Pa = 0.8 kPa$

Also $π = C R T = \frac{W_{solute}}{M_{solute} V_{solution}} \times R T$

$π = C R T$

$= \frac{20 g}{M_{solute} \times 1 {dm}^{3}} \times 8.3 {kPa dm}^{3} K^{- 1} {mol}^{- 1} \times 300 K$

$0.8 kPa = \frac{20 g}{M_{solute} \times 1 {dm}^{3}} \times 8.3 {kPa dm}^{3} K^{- 1} {mol}^{- 1} \times 300 K$

$M_{solute} = \frac{20 g}{0.8 kPa \times 1 {dm}^{3}} \times 8.3 {kPa dm}^{3} K^{- 1} {mol}^{- 1} \times 300 K$

$= 62250 {g mol}^{- 1} = 62.25 {kg mol}^{- 1}$

Topic Question Set

Q 11 :

The vapour pressure vs. temperature curve for a solution–solvent system is shown below:

The boiling point of the solvent is ______ °C. [2023]

Q 13 :

An aqueous solution of volume 300 ${cm}^{3}$ contains 0.63 g of protein. The osmotic pressure of the solution at 300 K is 1.29 mbar. The molar mass of the protein is ________ g ${mol}^{- 1}$ . Given: R = 0.083 L bar $K^{- 1}$ ${mol}^{- 1}$ [2023]

Q 14 :

Solutions of 12 g of non-electrolyte (A) prepared by dissolving it in 1000 mL of water exerts the same osmotic pressure as that of 0.05 M glucose solution at the same temperature. The empirical formula of A is ${CH}_{2} O$ . The molecular mass of A is ______ g. (Nearest integer) [2023]

Q 20 :

A sample of sea water was found to contain $9.5 %$ , w/w of $M g C l_{2}$ and $5.85 %$ , w/w of NaCl (Assuming complete ionization of salts) What will be the boiling point of sample?

( $K_{b}$ for water $= 0.52$ ,K-kg/mol, At. Wt of Mg = 24, Na = 23, Cl = 35.5)

Q 21 :

Molar depression constant for a solvent is $8.0 K kg {mol}^{- 1}$ . The depression in freezing point of the solvent for $0.03 mol {kg}^{- 1}$ solution of $K_{2} S O_{4}$ is ______ ?

(Assume complete dissociation of the electrolyte)

Q 22 :

The freezing point of equimolal aqueous solutions will be highest for:

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Topic Question Set

Q 11 : The vapour pressure vs. temperature curve for a solution–solvent system is shown below: The boiling point of the solvent is ______ °C. [2023]

Q 13 : An aqueous solution of volume 300 cm3 contains 0.63 g of protein. The osmotic pressure of the solution at 300 K is 1.29 mbar. The molar mass of the protein is ________ g mol-1. Given: R = 0.083 L bar K-1 mol-1 [2023]

Q 14 : Solutions of 12 g of non-electrolyte (A) prepared by dissolving it in 1000 mL of water exerts the same osmotic pressure as that of 0.05 M glucose solution at the same temperature. The empirical formula of A is CH2O. The molecular mass of A is ______ g. (Nearest integer) [2023]

Q 20 : A sample of sea water was found to contain 9.5%, w/w of MgCl2 and 5.85%, w/w of NaCl (Assuming complete ionization of salts) What will be the boiling point of sample? ( Kb for water =0.52 ,K-kg/mol, At. Wt of Mg = 24, Na = 23, Cl = 35.5)

Q 21 : Molar depression constant for a solvent is 8.0 K kg mol-1. The depression in freezing point of the solvent for 0.03 mol kg-1 solution of K2SO4 is ______ ? (Assume complete dissociation of the electrolyte)

Q 22 : The freezing point of equimolal aqueous solutions will be highest for:

Q 24 : At 27°C, 0.1 M, 1 L K4[Fe(CN)6] aqueous solution and 0.1 M, 1 L FeCl3 aqueous solution are placed in a container separated by a semi permeable membrane AB. Assume complete dissociation of both the solutes. Which of the following statement is correct? [2026] [IMAGE 202]

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Q 11 :

The vapour pressure vs. temperature curve for a solution–solvent system is shown below:

The boiling point of the solvent is ______ °C. [2023]

Q 13 :

An aqueous solution of volume 300 ${cm}^{3}$ contains 0.63 g of protein. The osmotic pressure of the solution at 300 K is 1.29 mbar. The molar mass of the protein is ________ g ${mol}^{- 1}$ . Given: R = 0.083 L bar $K^{- 1}$ ${mol}^{- 1}$ [2023]

Q 14 :

Solutions of 12 g of non-electrolyte (A) prepared by dissolving it in 1000 mL of water exerts the same osmotic pressure as that of 0.05 M glucose solution at the same temperature. The empirical formula of A is ${CH}_{2} O$ . The molecular mass of A is ______ g. (Nearest integer) [2023]

Q 20 :

A sample of sea water was found to contain $9.5 %$ , w/w of $M g C l_{2}$ and $5.85 %$ , w/w of NaCl (Assuming complete ionization of salts) What will be the boiling point of sample?

( $K_{b}$ for water $= 0.52$ ,K-kg/mol, At. Wt of Mg = 24, Na = 23, Cl = 35.5)

Q 21 :

Molar depression constant for a solvent is $8.0 K kg {mol}^{- 1}$ . The depression in freezing point of the solvent for $0.03 mol {kg}^{- 1}$ solution of $K_{2} S O_{4}$ is ______ ?

(Assume complete dissociation of the electrolyte)

Q 22 :

The freezing point of equimolal aqueous solutions will be highest for: