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The vapour density of undecomposed N2O4 is 46. When heated, vapour density decreases to 24.5 due to its dissociation to NO2. The percent dissociation of N2O4 at the final temperature is: 

(a) 87                                         (b) 60

(c) 40                                         (d) 70

Concept Videos :-

#6 | Van't Hoff Factor & Abnormal Molecular Mass

Concept Questions :-

Van’t Hoff Factor

(a) normal molar mass/exp. molar mass = 1+ α; (Molar mass = 2 x VD)

M1 =2*46=92

m2=2*24.5=49

                                                  92/49 = 1+ α

                                                ..α =0.87

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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

Concept Videos :-

Concept Questions :-

Dalton’s Law of Partial Pressure/ Henry"s law

(b) 15.03 cmCH4(NTP) is dissolved in 500 cm3 water.

Thus, we have Volume v dissolved in 1 cm3 water = 15.30500×1=0.03006cm3

Amount of gas dissolved = PVRT=1×0.0300682.06×273

                                    =1.34 X 10-6mol

Mass of gas disolved, m=1.34 X  10-6 X 16 = 2.144 X  10-5gm

Now, according to Henry's law, we have m=Kp (K- Henry constant. K=m/p.

Substituting the values of m and p, we get K=2.144 X 105gm atm-1

Now for 1 X  10-3mole of gas in 300cm-3 of water we have Mass of gas dissolved in 1 cm3 water

=1×10-3×16300×1=5.33×10-5gm

From Henry's law, pressure required to dissolve the above amount = mk=5.33×10-52.144×10-5=2.486 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

Concept Videos :-

#10 | Elevation in Boiling Point

Concept Questions :-

Elevation of Boiling Point

(b) Designating water and benzene by 1 and 2

0.002T12l1=1.85, 0.002T22l2=5.125.121.85=T2T12×l1l2=T2T12×83T2T12=5.121.85×38T2=5.12×31.85×812×273=278.12K=5.12°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

Concept Videos :-

#9 | Osmosis & Osmotic Pressure

Concept Questions :-

Osmosis and Osmotic Pressure

(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 Ksp (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.

XY2(s)          X2+(aq) + 2Y-(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.

Concept Videos :-

#3 | Introduction to Colligative Properties
#4 | Relative Lowering Introduction to Colligative Properties

Concept Questions :-

Relative Lowering of Vapour Pressure/ Vapour Pressure

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

AB2(s)          A2+(aq) + 2B-(aq)

                       s              2s

P0-PsPs=nN; 31.82-31.7831.78=3s55.56

s=0.0233 mol/lit

Ksp (at 30°C) = (0.0233) (2×0.0233)2 =5.05 × 10-5(mol/lit)3Ksp (at 25°C) = 3.56 ×10-5(mol/lit)3Now, log Ksp30°CKsp25°C=H2.303RT2-T2T1T2          log 5.05 ×10-53.56 ×10-5=H2.303×8.3145303 ×298H = 52.5kJ/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

Concept Videos :-

#3 | Introduction to Colligative Properties
#4 | Relative Lowering Introduction to Colligative Properties

Concept Questions :-

Relative Lowering of Vapour Pressure/ Vapour Pressure

(b) PP0=XB, so XB=760-750760=176

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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

Concept Videos :-

#9 | Osmosis & Osmotic Pressure

Concept Questions :-

Osmosis and Osmotic Pressure

(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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The latent heat of vapourisation of water is 540 cal g-1 at 100°C. Kb for water is

(a) 0.562 K.mole-1

(b) 1.86 K.mole-1

(c) 0.515 K.mole-1

(d) 5.12 K.mole-1

 

Concept Videos :-

#10 | Elevation in Boiling Point

Concept Questions :-

Elevation of Boiling Point

(c)

Kb=0.002×TB2LV=0.002×(373)2540=0.515 K mole-1

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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

Concept Videos :-

#9 | Osmosis & Osmotic Pressure

Concept Questions :-

Osmosis and Osmotic Pressure

(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

Concept Videos :-

#11 | Depression in Freezing Point
#12 | Ques : Elevation in Boiling Point & Depression in Freezing Point

Concept Questions :-

Depression of Freezing Point

(b) Let the moles of benzene vapourizes at 20° =n1Volume on n1 mole of benzene (l) = 78n10.877Volume of n1 mole of benzene (g) = 275078n10.877PV = nRTP78n10.877 27501000 = n1 ×0.082 ×293PBenzene0=0.0982 atm = 74.63 mmHgPBenzene0 at 27°C can be calculated as logP2P1 = Hvap2.303 RT2-T1T1T2logP274.63 = 394.57 × 782.303 × 8.3147300 × 293PBenzene0 (at 27°C) = 100.2mm HgMolality of the solution = Xsolute ×1000Xsolvent × 78=100.2 - 98.8898.88×10000.98 × 78=0.17Tf=Kfm = 5 ×0.17 = 0.85We know that Kf = RTf21000 HfM5=8.134 × Tf21000×1006078Tf =278.5K Freezing point of solution = 278.5 - 0.85= 277.65 K

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