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Chemistry NEET Questions — Free MCQ Practice

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At 298 K, 500cm³ H₂O dissolved 15.30 cm³ CH₄(STP) under a partial pressure of methane of one atm. If Henery's law holds, what pressure is required to cause 0.001 mole methane to dissolve in 300cm³ water ?

A.

0.286 atm

B.

2.486 atm

C.

1.286 atm

D.

3.111 atm

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The molal depression constant for water=1.85 deg/molal and for benzene is 5.12 deg/molal. If the ratio of the latent heats of fusion of benzene to water is 3:8, calculate the freezing point of benzene.

A.

5.75 $° C$

B.

5.12 $° C$

C.

4.97 $° C$

D.

6.1 $° C$

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Which of the following colligative properties is associated with the concentration term 'molarity'?

A.

Lowering of vap.pressure

B.

Osmotic pressure

C.

Depression in freezing point

D.

Elevation in boiling point

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Explanation

(b) p=CRT where c is the molarity of the solution. Hence, osmotic pressure is only associated with the molarity of the solution.

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The K_sp (25 $° C$ ) of sparingly soluble salt XY₂(s) is 3.56X10^-5(mol lit^-1)³ and at 30 $° C$ the vapour pressure of its saturated solution in water is 31.78 mm Hg. Calculate the enthalpy change of the reaction.

${XY}_{2} (s) \overset{}{⇌} X^{2 +} (aq) + 2 Y^{-} (aq)$

Given: Vapour pressure of pure water=31.82 mm Hg

A.

42.5 kJ/mol.

B.

32.1 kJ/mol.

C.

24.5 kJ/mol.

D.

52.5 kJ/mol.

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Explanation

(d) Let the solublity of AB₂(s) at 30 $° C$ be 's' mol Lit^-1

${AB}_{2} (s) \overset{}{⇌} A^{2 +} (aq) + 2 B^{-} (aq)$

s 2s

$\frac{P^{0} - P_{s}}{P_{s}} = \frac{n}{N}; \frac{31.82 - 31.78}{31.78} = \frac{3 s}{55.56}$

s=0.0233 mol/lit

$K_{sp} (at 30 ° C) = (0.0233) (2 \times 0.0233)^{2} = 5.05 \times 10^{- 5} (mol / lit)^{3} K_{sp} (at 25 ° C) = 3.56 \times 10^{- 5} (mol / lit)^{3} Now, \log \frac{K_{sp} (30 ° C)}{K_{sp} (25 ° C)} = \frac{∆ H}{2.303 R} [\frac{T_{2} - T_{2}}{T_{1} T_{2}}] \log (\frac{5.05 \times 10^{- 5}}{3.56 \times 10^{- 5}}) = \frac{∆ H}{2.303 \times 8.314} (\frac{5}{303 \times 298}) ∆ H = 52.5 kJ / mol$

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The vapoure pressure of a dilute aqueous solution of glucose is 750mm of mercury at 373K. The mole fraction of solute in the solution is-

A.

1/10

B.

1/76

C.

1/7.6

D.

1/35

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Explanation

(b) $\frac{∆ P}{P_{0}} = X_{B}, so X_{B} = \frac{760 - 750}{760} = \frac{1}{76}$

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Which one of the folowing pairs of solution can we expect to be isotonic at the same temperature-

A.

0.1 (M) urea and 0.1 (M) NaCl

B.

0.1 (M) urea and 0.2 (M) MgCl₂

C.

0.1 (M) NaCl and 0.1 (M) Na₂SO₄

D.

0.1 (M) Ca(NO₃)₂ and 0.1 (M) Na₂SO₄

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Explanation

(d) As the no. of ionic specied produced after complete dissociation of 0.1 (M) Ca(NO₃)₂ and 0.1(M) Na₂SO₄ are same

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Which of the following aqueous solution has osmotic pressure nearest to that an equimolar solution of K₄[Fe(CN)₆]?

A.

Na₂SO₄

B.

BaCl₂

C.

Al₂(SO₄)₃

D.

C₁₂H₂₂O₁₁

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Explanation

(c) As the no. of species obtained after the complete dissociation of Al₂(SO₄)₃ is same to the no. of species obtained after complete dissociation of K₄[Fe(CN)₆]

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The molar volume of liquid benzene (density= 0.877 g ml^-1) increases by a factor of 2750 as it vapourises at 20 $° C$ . At 27 $° C$ when a non-volatile solute (that does not dissociate) is dissolved in 54.6cm3 of benzene, vapour pressure of this solution, is found to be 98.88 mm Hg. calculate the freezing point of the solution.

Given: Enthalpy of vapourisation of benzene(l)=394.57 Jg^-1. Enthalpy of fusion of benzene (l) = 10.06 kJ mol^-1 Molal depression constant for benzene=5.0 K kg mol^-1.

A.

177.65 K

B.

277.65 K

C.

517.65 K

D.

237.15 K

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Explanation

$(b) Let the moles of benzene vapourizes at 20 ° = n_{1} Volume on n_{1} mole of benzene (l) = \frac{78 n_{1}}{0.877} Volume of n_{1} mole of benzene (g) = 2750 (\frac{78 n_{1}}{0.877}) PV = nRT P (\frac{78 n_{1}}{0.877}) (\frac{2750}{1000}) = n_{1} \times 0.082 \times 293 P_{Benzene}^{0} = 0.0982 atm = 74.63 mmHg P_{Benzene}^{0} at 27 ° C can be calculated as \log (\frac{P_{2}}{P_{1}}) = \frac{∆ H_{vap}}{2.303 R} [\frac{T_{2} - T_{1}}{T_{1} T_{2}}] \log (\frac{P_{2}}{74.63}) = \frac{394.57 \times 78}{2.303 \times 8.314} [\frac{7}{300 \times 293}] P_{Benzene}^{0} (at 27 ° C) = 100.2 mm Hg Molality of the solution = \frac{X_{solute} \times 1000}{X_{solvent} \times 78} = \frac{(\frac{100.2 - 98.88}{98.88}) \times 1000}{0.98 \times 78} = 0.17 ∆ T_{f} = K_{f} m = 5 \times 0.17 = 0.85 We know that K_{f} = [\frac{{RT}_{f}^{2}}{1000 ∆ H_{f}}] M 5 = (\frac{8.134 \times T_{f}^{2}}{1000 \times 10060}) 78 T_{f} = 278.5 K ∴ Freezing point of solution = 278.5 - 0.85 = 277.65 K$

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Calculate the depression in the freezing point of water when 10 g of CH₃CH₂CHClCOOH is added to 250 g of water.

K_a=1.4 X 10-³, K_f=1.86K kg mol^-1

A.

0.15 $° C$

B.

2.15 $° C$

C.

1.15 $° C$

D.

0.65 $° C$

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Explanation

(d)

Mass of the acid = 10 g

Mass of water = 250 g

Mass pf acid per kg of water = $\frac{10 g \times 1000 g}{250 g} = 40 g$

Molar mass of the acid = 122.5 g/mol

The given acid is a weak acid. So,

$α = \sqrt{\frac{K_{a}}{C}} = \sqrt{\frac{1.4 \times 10^{- 3}}{(\frac{40}{122.5})}} = 0.065 So, i = α (n - 1) + 1 = 0.065 (2 - 1) + 1 = 1.065 Then, ∆ T_{f} = {iK}_{f} m = 1.065 \times 1.86 \times \frac{40}{122.5} = 0.65 K SO, the depression in the freezing of water will be 0.65 ° C .$

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The ratio of the vapour pressure of a solution to the vapour pressure of the solvent is

A.

Equal to the mole fraction of the solvent

B.

Proportional to the mole fraction of the solute

C.

Equal to the mole fraction of the solute

D.

None of the above

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Explanation

(a) This is one of forms of statements of Raoult's law

$\frac{P - P_{s}}{P} = \frac{n}{n + N}$

Where n=number of moles of solute, P and P_s are the vapour pressure of the solvent and solution.

Chemistry MCQs for NEET — Practice Questions with Answers

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