All questions

• A wire of resistance R is stretched uniformly until its length is doubled. Its new resistance is Answer: C. 4R Volume constant: A→ A/2 . R=ρl/A→ρ2l/A/2=4R . Electric Current Through Conductors 1 answer
• The packing efficiency of a body-centred cubic (bcc) unit cell is Answer: C. 68% In bcc, 4r=sqrt3a ; two atoms occupy 2·4/3π r³/a³≈0.68 of the cell. Solid State 1 answer
• An element with molar mass 60 g mol-1 forms an fcc lattice with edge length 400 pm. Its density is approximately (\(N_A=6.022\times10^{23}\)) Answer: D. 6.23 g cm-3 ρ=ZM/a³N_A=frac4×60(4×10⁻⁸)³×6.022×10²³=240/38.54≈6.23 g cm⁻³. Solid State 1 answer
• In a cubic solid, atoms A occupy the corners and atoms B occupy the face centres. If one A atom is missing from one corner of each unit cell, the formula of the compound is Answer: D. A7B24 A =7×1/8=7/8 , B =6×1/2=3 . Ratio 7/8:3=7:24 . Solid State 1 answer
• Which of the following defects does NOT change the density of an ionic crystal? Answer: D. Frenkel defect In a Frenkel defect an ion only shifts to an interstitial site, so mass and volume stay the same. Solid State 1 answer
• 1.8 g of glucose (molar mass 180 g mol-1) is dissolved in 100 g of water. The elevation in boiling point is (\(K_b=0.52\) K kg mol-1) Answer: D. 0.052 K m=0.01/0.1=0.1 mol kg⁻¹; Δ T_b=0.52×0.1=0.052 K. Solutions 1 answer
• The osmotic pressure of a 0.1 M aqueous solution of a non-electrolyte at 300 K is (R = 0.082 L atm K-1 mol-1) Answer: C. 2.46 atm π=CRT=0.1×0.082×300=2.46 atm. Solutions 1 answer
• Which of the following liquid pairs shows a negative deviation from Raoult's law? Answer: C. Acetone + chloroform Acetone and chloroform form hydrogen bonds with each other, so the A–B attraction is stronger than A–A and B... Solutions 1 answer
• The pH of 0.1 M acetic acid (\(K_a=1.8\times10^{-5}\)) is approximately Answer: B. 2.87 [H^+]=√(K_aC)=sqrt1.8×10⁻⁶=1.34×10⁻³ M; pH ≈2.87 . Ionic Equilibria 1 answer
• The solubility of Ag2CrO4 in water is \(1\times10^{-4}\) mol L-1. Its solubility product is Answer: A. 4×10⁻¹² K_sp=(2s)²s=4s³=4×10⁻¹² . Ionic Equilibria 1 answer
• A buffer contains 0.02 M acetic acid and 0.2 M sodium acetate. If p\(K_a\) of acetic acid is 4.74, the pH of the buffer is Answer: B. 5.74 pH =pK_a+log[salt]/[acid]=4.74+log10=5.74 . Ionic Equilibria 1 answer
• For the reaction \(\mathrm{N_2(g)+3H_2(g)\rightarrow2NH_3(g)}\) at 298 K, \(\Delta H-\Delta U\) is approximately Answer: A. -4.96 kJ Δ n_g=2-4=-2 ; Δ H-Δ U=Δ n_gRT=-2×8.314×298≈-4955 J. Chemical Thermodynamics 1 answer
• For a reaction \(\Delta H=+30\) kJ mol-1 and \(\Delta S=+100\) J K-1 mol-1. The reaction becomes spontaneous above Answer: B. 300 K Δ G<0 when T>Δ H/Δ S=30000/100=300 K. Chemical Thermodynamics 1 answer
• Given: \(\mathrm{C(s)+O_2(g)\rightarrow CO_2(g)}\), \(\Delta H=-393.5\) kJ; \(\mathrm{CO(g)+\frac12O_2(g)\rightarrow CO_2(g)}\), \(\Delta H=-283.0\) kJ. The standard enthalpy of formation of CO(g) is Answer: D. -110.5 kJ mol-1 Subtract the second equation from the first: Δ_fH(CO)=-393.5-(-283.0)=-110.5 kJ mol⁻¹. Chemical Thermodynamics 1 answer
• For the Daniell cell \(\mathrm{Zn|Zn^{2+}(1M)||Cu^{2+}(1M)|Cu}\), \(E^\circ_{\mathrm{Zn}}=-0.76\) V and \(E^\circ_{\mathrm{Cu}}=+0.34\) V. The standard Gibbs energy change is Answer: C. -212.3 kJ E^°_cell=0.34-(-0.76)=1.10 V; Δ G^°=-nFE^°=-2×96500×1.10=-212.3 kJ. Electrochemistry 1 answer
• A current of 2 A is passed through CuSO4 solution for 965 s. The mass of copper deposited is (Cu = 63.5, F = 96500 C) Answer: A. 0.635 g Q=1930 C =0.02 F; Cu²⁺ needs 2F/mol, so 0.01 mol =0.635 g. Electrochemistry 1 answer
• The electrode potential of a hydrogen electrode (\(P_{H_2}\) = 1 bar) dipped in a solution of pH 3 at 298 K is Answer: D. -0.177 V E=-0.0592 pH=-0.0592×3≈-0.177 V. Electrochemistry 1 answer
• The conductivity of a 0.1 M solution of an electrolyte is 0.0129 S cm-1. Its molar conductivity is Answer: A. 129 S cm2 mol-1 Lambda_m=1000kappa/C=1000×0.0129/0.1=129 S cm² mol⁻¹. Electrochemistry 1 answer
• The rate constant of a first-order reaction is \(6.93\times10^{-3}\) s-1. The time required for 75% completion is Answer: A. 200 s t_(1/2)=0.693/k=100 s; 75% completion takes two half-lives, i.e. 200 s. Chemical Kinetics 1 answer
• For the rate law \(r=k[\mathrm A]^2[\mathrm B]\), if [A] is doubled and [B] is halved, the rate becomes Answer: C. 2 times r'=k(2[mathrm A])²[mathrm B]/2=2r . Chemical Kinetics 1 answer
• The rate of a reaction doubles when temperature is raised from 300 K to 310 K. The activation energy is approximately (R = 8.314 J K-1 mol-1, log 2 = 0.301) Answer: C. 53.6 kJ mol-1 log2=E_a/2.303R·T₂-T₁/T₁T₂⇒ E_a=2.303×8.314×0.301×300×310/10≈53.6 kJ mol⁻¹. Chemical Kinetics 1 answer
• The shape of the XeF4 molecule is Answer: A. square planar Xe has 4 bond pairs and 2 lone pairs (sp³d²). The lone pairs sit trans to each other, giving a square planar shape. Elements of Groups 16, 17 and 18 1 answer
• The oxidation state of sulphur in peroxodisulphuric acid (H2S2O8) is Answer: A. +6 There is one peroxide O–O link (each O −1) and six oxygens at −2: 2+2x-2-12=0⇒ x=+6 . Elements of Groups 16, 17 and 18 1 answer
• The spin-only magnetic moment of Fe2+ ion (Z of Fe = 26) is Answer: A. 4.90 BM Fe²⁺ is 3d⁶ with 4 unpaired electrons: μ=√(4×6)=√(24)≈4.90 BM. Transition and Inner Transition Elements 1 answer