JEE Main Practice — Chemistry Set 6

JEE Main · 2026 · with answers and explanations

questions
19
Total marks
76
Duration
46 minutes
Negative marking
−1

Subjects: Chemistry

Question 1

Chemistry · 4 marks · −1

The limiting molar conductivities (S cm2 mol−1\mathrm{S\,cm^2\,mol^{-1}}) of HCl, NaCl and CH3COONa\mathrm{CH_3COONa} are 426, 126 and 91 respectively. A 0.001 M acetic acid solution has a molar conductivity of 48.9 S cm2 mol−148.9\ \mathrm{S\,cm^2\,mol^{-1}}. The dissociation constant of acetic acid is approximately

  1. A.1.8×10−51.8\times10^{-5}
  2. B.1.25×10−41.25\times10^{-4}
  3. C.3.9×10−43.9\times10^{-4}
  4. D.5.6×10−65.6\times10^{-6}

Correct answer: A

Explanation: Kohlrausch: Λm∘(CH3COOH)=91+426−126=391\Lambda^\circ_m(\mathrm{CH_3COOH})=91+426-126=391. α=48.9391=0.125\alpha=\frac{48.9}{391}=0.125. Ka=cα21−α=10−3×0.0156250.875≈1.8×10−5K_a=\frac{c\alpha^2}{1-\alpha}=\frac{10^{-3}\times0.015625}{0.875}\approx1.8\times10^{-5}.

Question 2

Chemistry · 4 marks · −1

The current (in ampere) that must be passed for 965 s through a CuSO4\mathrm{CuSO_4} solution to deposit 0.635 g of copper is ______. (Cu = 63.5, F=96500F=96500 C mol−1^{-1})

Correct answer: 2

Explanation: 0.635 g = 0.01 mol Cu needs 0.02 mol e−^- =1930=1930 C. I=1930965=2I=\frac{1930}{965}=2 A.

Question 3

Chemistry · 4 marks · −1

The conductivity of a 0.05 M solution of an electrolyte is 0.0065 S cm−10.0065\ \mathrm{S\,cm^{-1}}. Its molar conductivity (in S cm2 mol−1\mathrm{S\,cm^2\,mol^{-1}}) is ______.

Correct answer: 130

Explanation: Λm=1000κc=1000×0.00650.05=130 S cm2 mol−1\Lambda_m=\frac{1000\kappa}{c}=\frac{1000\times0.0065}{0.05}=130\ \mathrm{S\,cm^2\,mol^{-1}}.

Question 4

Chemistry · 4 marks · −1

4 g of NaOH is dissolved in 500 g of water. The molality of the solution is (Na = 23, O = 16, H = 1)

  1. A.0.1 m
  2. B.0.2 m
  3. C.0.008 m
  4. D.2 m

Correct answer: B

Explanation: Moles of NaOH =440=0.1=\frac{4}{40}=0.1. Molality =0.10.5 kg=0.2=\frac{0.1}{0.5\ \mathrm{kg}}=0.2 m.

Question 5

Chemistry · 4 marks · −1

The van't Hoff factor of K2SO4\mathrm{K_2SO_4} assuming complete dissociation in water is

  1. A.1
  2. B.2
  3. C.4
  4. D.3

Correct answer: D

Explanation: K2SO4→2K++SO42−\mathrm{K_2SO_4\to2K^++SO_4^{2-}}: 3 ions, so i=3i=3.

Question 6

Chemistry · 4 marks · −1

Liquids A (PA∘=300P_A^\circ=300 mm Hg) and B (PB∘=100P_B^\circ=100 mm Hg) form an ideal solution in which the mole fraction of A is 0.4. The mole fraction of A in the vapour in equilibrium with the solution is

  1. A.0.4
  2. B.23\frac23
  3. C.13\frac13
  4. D.0.6

Correct answer: B

Explanation: pA=0.4×300=120p_A=0.4\times300=120, pB=0.6×100=60p_B=0.6\times100=60, total 180 mm Hg. yA=120180=23y_A=\frac{120}{180}=\frac23 (vapour is richer in the more volatile component).

Question 7

Chemistry · 4 marks · −1

1.8 g of glucose (M=180M=180) is dissolved in 100 g of water. The depression in freezing point is (KfK_f of water =1.86=1.86 K kg mol−1^{-1})

  1. A.1.86 K
  2. B.0.0186 K
  3. C.0.186 K
  4. D.0.372 K

Correct answer: C

Explanation: m=0.010.1=0.1m=\frac{0.01}{0.1}=0.1 mol kg−1^{-1}. ΔTf=Kfm=1.86×0.1=0.186\Delta T_f=K_fm=1.86\times0.1=0.186 K.

Question 8

Chemistry · 4 marks · −1

A 5% (w/v) aqueous glucose solution is isotonic at the same temperature with a urea solution of concentration (glucose 180, urea 60 g mol−1^{-1}; both non-electrolytes)

  1. A.1.67% (w/v)
  2. B.5% (w/v)
  3. C.15% (w/v)
  4. D.3.33% (w/v)

Correct answer: A

Explanation: Isotonic solutions of non-electrolytes have equal molarity. Glucose: 50180\frac{50}{180} mol/L. Urea needed: 50180×60=16.7\frac{50}{180}\times60=16.7 g/L =1.67%=1.67\% (w/v).

Question 9

Chemistry · 4 marks · −1

1.22 g of benzoic acid (M=122M=122) dissolved in 20 g of benzene lowers its freezing point by 1.536 K (KfK_f of benzene =5.12=5.12 K kg mol−1^{-1}). Benzoic acid dimerises in benzene. The percentage of benzoic acid molecules that are associated into dimers is

  1. A.60%
  2. B.40%
  3. C.80%
  4. D.90%

Correct answer: C

Explanation: Molality (as monomer) =0.010.020=0.5=\frac{0.01}{0.020}=0.5. i=ΔTfKfm=1.5362.56=0.6i=\frac{\Delta T_f}{K_fm}=\frac{1.536}{2.56}=0.6. For 2A⇌A22A\rightleftharpoons A_2 with degree of association β\beta: i=1−β2=0.6⇒β=0.8i=1-\frac\beta2=0.6\Rightarrow\beta=0.8, i.e. 80%.

Question 10

Chemistry · 4 marks · −1

Commercial sulfuric acid is 98% H2SO4\mathrm{H_2SO_4} by mass and has density 1.8 g mL−1^{-1}. Its molarity is ______. (H2SO4=98\mathrm{H_2SO_4}=98 g mol−1^{-1})

Correct answer: 18

Explanation: 1 L has mass 1800 g, of which 0.98×1800=17640.98\times1800=1764 g is H2SO4=176498=18\mathrm{H_2SO_4}=\frac{1764}{98}=18 mol. Molarity =18=18 M.

Question 11

Chemistry · 4 marks · −1

A solution of 3 g of a non-volatile, non-electrolyte solute in 100 g of water freezes at −0.31∘-0.31^\circC. The molar mass of the solute (in g mol−1^{-1}) is ______. (Kf=1.86K_f=1.86 K kg mol−1^{-1})

Correct answer: 180

Explanation: ΔTf=0.31\Delta T_f=0.31 K =Kfw×1000M×W⇒M=1.86×3×10000.31×100=180=K_f\frac{w\times1000}{M\times W}\Rightarrow M=\frac{1.86\times3\times1000}{0.31\times100}=180 g mol−1^{-1}.

Question 12

Chemistry · 4 marks · −1

The number of molecules in 11 g of carbon dioxide is (NA=6.022×1023N_A=6.022\times10^{23} mol−1^{-1})

  1. A.6.022×10236.022\times10^{23}
  2. B.3.011×10233.011\times10^{23}
  3. C.1.505×10231.505\times10^{23}
  4. D.4.515×10234.515\times10^{23}

Correct answer: C

Explanation: Moles =1144=0.25=\frac{11}{44}=0.25. Molecules =0.25×6.022×1023=1.505×1023=0.25\times6.022\times10^{23}=1.505\times10^{23}.

Question 13

Chemistry · 4 marks · −1

5.85 g of NaCl is dissolved in water to make 500 mL of solution. The molarity of the solution is (Na = 23, Cl = 35.5)

  1. A.0.2 M
  2. B.0.1 M
  3. C.0.05 M
  4. D.2 M

Correct answer: A

Explanation: Moles =5.8558.5=0.1=\frac{5.85}{58.5}=0.1. Molarity =0.10.5=0.2=\frac{0.1}{0.5}=0.2 M.

Question 14

Chemistry · 4 marks · −1

3 g of hydrogen is ignited with 32 g of oxygen. The mass of water formed is

  1. A.36 g
  2. B.18 g
  3. C.27 g
  4. D.35 g

Correct answer: C

Explanation: 2H2+O2→2H2O\mathrm{2H_2+O_2\to2H_2O}. H2=1.5\mathrm{H_2}=1.5 mol needs 0.75 mol O2\mathrm{O_2}; 1 mol O2\mathrm{O_2} is available, so H2\mathrm{H_2} is limiting. Water =1.5=1.5 mol =27=27 g (0.25 mol O2\mathrm{O_2} is left over).

Question 15

Chemistry · 4 marks · −1

An organic compound contains 40.0% C, 6.67% H and the rest oxygen by mass. Its molar mass is 180 g mol−1^{-1}. Its molecular formula is

  1. A.C6H12O6\mathrm{C_6H_{12}O_6}
  2. B.CH2O\mathrm{CH_2O}
  3. C.C5H10O5\mathrm{C_5H_{10}O_5}
  4. D.C6H6O6\mathrm{C_6H_6O_6}

Correct answer: A

Explanation: Mole ratio C : H : O =4012:6.671:53.3316=3.33:6.67:3.33=1:2:1=\frac{40}{12}:\frac{6.67}{1}:\frac{53.33}{16}=3.33:6.67:3.33=1:2:1. Empirical formula CH2O\mathrm{CH_2O} (30). n=18030=6n=\frac{180}{30}=6: C6H12O6\mathrm{C_6H_{12}O_6}.

Question 16

Chemistry · 4 marks · −1

The mole fraction of the solute in a 1 molal aqueous solution is approximately

  1. A.0.1
  2. B.0.0177
  3. C.0.018
  4. D.0.0555

Correct answer: B

Explanation: 1 mol solute in 1000 g water =55.56=55.56 mol water. x=11+55.56≈0.0177x=\frac{1}{1+55.56}\approx0.0177.

Question 17

Chemistry · 4 marks · −1

20 mL of a gaseous hydrocarbon is exploded with excess oxygen. After cooling to room temperature (water condenses), the gas volume has contracted by 50 mL. On passing the remaining gas through aqueous KOH, a further contraction of 60 mL occurs. All volumes are at the same temperature and pressure. The hydrocarbon is

  1. A.C3H8\mathrm{C_3H_8}
  2. B.C2H6\mathrm{C_2H_6}
  3. C.C3H4\mathrm{C_3H_4}
  4. D.C3H6\mathrm{C_3H_6}

Correct answer: D

Explanation: CxHy+(x+y4)O2→xCO2+y2H2O(l)\mathrm{C_xH_y+(x+\frac y4)O_2\to xCO_2+\frac y2H_2O(l)}. KOH absorbs CO2\mathrm{CO_2}: 20x=60⇒x=320x=60\Rightarrow x=3. Contraction on cooling =V(1+x+y4)−Vx=20(1+y4)=50⇒y=6=V\left(1+x+\frac y4\right)-Vx=20\left(1+\frac y4\right)=50\Rightarrow y=6. Hydrocarbon: C3H6\mathrm{C_3H_6}.

Question 18

Chemistry · 4 marks · −1

The volume (in mL) of 0.1 M NaOH required to neutralise 25 mL of 0.08 M H2SO4\mathrm{H_2SO_4} completely is ______.

Correct answer: 40

Explanation: H+\mathrm{H^+} from acid =25×0.08×2=4=25\times0.08\times2=4 mmol. NaOH needed =4=4 mmol =40.1=40=\frac{4}{0.1}=40 mL.

Question 19

Chemistry · 4 marks · −1

10 g of an impure sample of calcium carbonate gives 3.52 g of CO2\mathrm{CO_2} on complete decomposition (impurities do not decompose). The percentage purity of the sample is ______.

Correct answer: 80

Explanation: CaCO3→CaO+CO2\mathrm{CaCO_3\to CaO+CO_2}. CO2=3.5244=0.08\mathrm{CO_2}=\frac{3.52}{44}=0.08 mol, so CaCO3=0.08×100=8\mathrm{CaCO_3}=0.08\times100=8 g. Purity =810×100=80%=\frac{8}{10}\times100=80\%.

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