Title Chemistry ÔÇó Final Step-A ÔÇó SOL.pdf ✅

Vidyamandir Classes VMC | Final Step-A 1 Class XI | Chemistry Solutions to Final Step-A | Chemistry Stoichiometry - 1 & Redox Reactions 1.(D) As H2O2 is loosing e s  2.(D) 1 1 5 HClO HCl HClO3      3.(A) 4.(D) [Cr (NH3)4Cl2] + : Cr 3   5.(C) 7 5e 2 MnO Mn 4      ; 7 6 1e 2 MnO Mn O 4 4       ; 7 4 3e MnO Mn O 4 2      ; 7 3 4e MnO Mn O 4 2 3      6.(C) Anode - where oxidation takes place 7.(C) 9 T( F) 32 T( C) 5     8.(D) He 1mole  ; Na 2 moles  ; 2 Ca 10 moles   ; He 3moles  9.(A)   3 H SO moles 0.1 0.02 2 10 2 4       2  10–3 (H2SO4)  2  10–3  NA molecules Moles of = 12.044 × 1020 molecules 10.(A)   g 3.12 1.5  to be one decimal as minimum significant figure has one digit after decimal. 11.(C) Example:   2 2     2 g 1 H O g H O 2    12.(B) Typical case of Law of multiple proportions. 13.(AD) O 32g 2  For 1 mol there are NA molecules 14.(AB) 20 1000 5 [NaOH] M M 40 200 2     ;   1000 5 KOH M 0.5 M 200 2     15.(CD) Refer formula 16.(BD) Refer theory 17.(A) As bromine case more change in oxidation state 18.(A) NaH Oxidation State 1     as compared to iodine 19.(B)  NH NO 4 3    3 5       20.(D)   5 2 Mn 3d 4s 7   21.(D) 22.(ABC) 23.(CD) 24.(CD) C : Vanadium, D : Phosphorous 25.(B) As oxidizing nature depends on electron gain enthalpy and hydration energy. 26.(C) Mn acts as oxidizing agent and oxidation state changes from +7 to +2.
Vidyamandir Classes VMC | Final Step-A 2 Class XI | Chemistry 27.(D)    y/ 2  2 2 2 V Vx V y y CxHy x O CO H O 4 2            H O CO 2 2 0.72 3.08 n n 18 44                2 2 CO H O n 3.08 18 n 44 0.72    x 7 y / 2 4   x 7 y 8  28.(D) Total Mass 1120 Volume ml 1.15                120 1000 1.15 M 2.05M 60 1120           29.(C)   g 0.75 0.6 1 g 22.05g 294 / 6      30.(C) As Cl reduces Mn to Mn2+ 31.(D) At STP 1 mole H 22.4L 2   6 mole HCl 3H  2  1 mole 2 1 HCl mole H 2       32.(B) Mass will remain constant 33.(B) 3 2 3 BCl H B 3HCl 2    2 moles of B 3 moles of H  2  34.(A) Mole ratio : 9 1 3.5 : : 3 : 4 : 1 12 1 14  . 35.(D) 36.(C) 37.(C) Refer structure from module 38.(B) 3 0 3 NH NO N 2H O 4 2 2 2     39.(C) meq FeC O meq K C O 2 4 2 2 7  3 × 1 = 6  mmoles of K2Cr2O7 40.(ABC)   2 4 0.1 NaHC O 0.1 moles of H   geq of 2 C O 2 0.1 0.2 2 4     41.(B) 16.9% solution of AgNO3 means 16.9 g of AgNO3 in 100 mL of solution. 16.9% of AgNO3 in 100 mL solution  8.45g of AgNO3 in 50 mL 0.049mol  Similarly, 5.8% of NaCl 2.9 g of NaCl in 50 mL solution  0.05 mol. The reaction can be represented as : 3 3 0.049 0 0.05 0 AgNO NaCl AgCl NaNO Initially    Finally 0 0 0.049 mol 0.049 mol  Mass of AgCl precipitated = 0.049 × 143.3  7.0g 42.(A) Mass of 1 mol (6.022 × 1023 atoms) of carbon = 12 g. If Avogadro number is changed to 6.022 × 1020 atoms then mass of 1 mol of carbon 20 3 23 12 6.022 10 12 10 g 6.022 10       
Vidyamandir Classes VMC | Final Step-A 3 Class XI | Chemistry 43.(C) According to Avogadro’s hypothesis, ratio of the volumes of gases will be equal to the ratio of their moles. H O CH 2 2 4 g g g n : n ; n 2 32 16    So, the ratio is g g g : : or 16 :1 : 2 2 32 16 . 44.(D) Moles of urea = 20 23 6.02 10 0.001 6.02 10    Concentration of solution = 0.001 1000 0.01M 100   45.(B) 2 2 2 1 H O H O 2   1 mol 0.5 mol 1 mol 10 g of H2 = 5 mol and 64 g of O2 = 2 mol  In this reaction, oxygen is the limiting reagent hence amount of H2O produced depends on that of O2. Since 0.5 mol of O2 gives 1 mol H2O  2 mol of O2 will give 4 mol H2O 46.(D) Average isotopic mass of X = 200 90 199 8 202 2 90 8 2        = 18000 1592 404 199.96 amu ~ 200 amu 100    47.(A) At STP: 22.4 L H2 = 6.023 × 1023 molecules 23 23 2 6.023 10 15 15L H 4.033 10 22.4      23 23 2 6.023 10 5 5 L N 1.344 10 22.4      23 23 2 6.023 10 0.5 0.5 g H 1.505 10 2      23 23 2 6.023 10 10 10 g of O 1.882 10 32      48.(D) Zeros placed left to the number are never significant, therefore the number of significant figures for the numbers. 161 cm, 0.161 and 0.0161 are same, i.e. 3. 49.(A) Amount of iron in one molecule (in gms) = 67200 0.334 224.45 amu 100   Number of iron atoms in one molecule of hemoglobin = 224.45 4 56  . 50.(A) 4NH 5O 4NO 6H O 3(g) 2(g) (g) 2 ( )    l 4 mol NH 5 moles O 4 moles NO 6 moles H O 3(g) 2(g) (g) 2 ( )    l Observe that O2 is limiting reagent, so it will be consumed totally leaving behind NH3. 51.(A) 1 mol CCl4 vapour = 12 + 4 × 35.5 = 154 gm At S.T.P.: Density of CCl4 vapour = 154 1 g L 22.4  = 1 6.875 g L .
Vidyamandir Classes VMC | Final Step-A 4 Class XI | Chemistry 52.(A) 2 Mg Na CO MgCO 2Na 2 3 3      1 gmeq of Mg2+ requires 1 gmeq of soda. 3 3 12.00 10 12.00mg 12.00 10 g 1000 24 2        = 1 meq of Mg2+  1 meq of soda is required. 53.(C) Both FeCl2 and SnCl2 are reducing agents with lower oxidation numbers. 54.(C) H2O2 acts as reducing agent in all those reactions in which O2 is evolved.. 55.(C) 1 5 2 6 6 1 KClO (COOH) H SO K SO KCl CO H O 3 2 2 4 2 4 2 2             Maximum change in oxidation number of chlorine, i.e., from +5 to 1 . 56.(C) 2 4 2 5e MnO 8H Mn 4H O        ; 3 FeC O Fe 2CO 3e 2 4 2      [Both Fe2+ ion and 2 C O2 4  ions will be oxidized] gmeq of MnO4  = gmeq of FeC2O4  3 5n 3 1 n 0.6 5      57.(B) 0 1 1 H O Br HOBr HBr 2 2      In the above reaction the oxidation number of Br2 increases from zero (in Br2) to + 1 (in HOBr) and decreases from zero (in Br2) to 1 (in HBr). Thus Br2 is oxidised as well as reduced and hence it is a disproportionation redox reaction. 58.(B) meq of NaOCl = meq of NaCrO2. 2 OCl H 2e Cl 2OH ; CrO 4OH CrO 2H O 3e 2 2 4 2                            n-factor of NaOCl = 2 n-factor of NaCrO2 = 3  2 0.15 V 3 0.2 20 V 40mL        59.(D) meq of H2O2 = meq of MnO / H 4    2 2 1 1 N 10 10 N(H O ) 0.56 0.56      Volume strength of H2O2 ; 1 N H2O2 1 5.6 10vol 0.56    60.(C) Find oxidation state of each element. (A) 3 VO [V 5] 4    (B) Mn O [Mn 3] 2 3   (C) 2 S O [S 2.5] 4 6    (D) Cl O [Cl 7] 2 7  