Title Chem Final Step-C Solutions.pdf ✅

Vidyamandir Classes APP | Stoichiometry-I & II 1 Solutions | Chemistry Solutions - Advanced Problem Package | Chemistry Stoichiometry-I & II 1.(D) 2CrO5 +3H2SO4Cr2(SO4)3+ 3H2O + 7/2 O2. 6 3 0 1 1 2 2Cr 6e 2Cr 7O 7O 7e O e O                     2.(B) 2 2 4 2 4 2 2 4 mol (a) mol (b) Na C O KHC O .H C O  With NaOH : 2 2 4 2 4 2 2 4 4 eq of Na C O eq of KHC O H C O eq of KMnO   2a 4b 5 0.2 V     .......(i) With KMnO4 : 2 4 2 2 4 gmeq of KHC O .H C O (n 3 as acid) gmeq of NaOH   3b 1 0.2 V    .........(ii) Divide (i)/(ii): 2a 4b a 2b 5 5 3b 3b 2      a 11 b 2   3.(C) 2 2 4 4Au 16KCN 6H O 3O 4KAu(CN) 12KOH      . V(OH) Cl FeCl 2HCl VOCl 3H O FeCl 4 2 2 2 3      . 4 2 4 2 2 4KMnO 4KOH 4K MnO O 2H O.     In above reaction out of 4 mol of 2 O  present in KOH (R.A) , two are oxidised to O2 and other two moles remained as such and went to H2O. 2 3 4 3 2 2 2MnO 5PbO 10HNO 2HMnO 5Pb(NO ) 4H O.      4.(C) x 2 x XeF H Xe xHF 2    1000 m mol of Xe 22.4 1 22400    ; m mol of HF (n = 1) = 6  x = 6 5.(B) Ist Part meq of I2(n= 2) = meq of 2 S O2 3  (n = 1) = 15 × 0.4 ×1 = 6, m mol of I2=3 IInd Part meq of I2 reacted = meq of NaOH initially – meq of H2SO4 with NaOH = 100 × 0.3 – 2 × 10 × 0.3 = 24 m mol of I2 (n = 2) reacted with NaOH = 24 12 2  Total m mol of I2 = 3 +12 = 15 ; [I2] = 15 0.1 150 
Vidyamandir Classes APP | Stoichiometry-I & II 2 Solutions | Chemistry 6.(D) 2 n 1 n 1 HCN KOH KCN H O      m mol of HCN meq of HCN meq of KOH 100    2 4 3 2 n 10 n 5 V ? HCN KMnO Mn NO CO          As HCN used in same for KMnO4 as is used for KOH. m mol of HCN (n = 1) = 100. meq of HCN (n = 10) = 100 × 10 = 1000 meq of HCN = meq of KMnO4  1000 = 5 × 5 × V :   V 40 mL 7.(A) 6 2 Mn O4   on disproportination gives MnO & MnO 4 2  2 4 4 [MnO MnO e ] 2       2 MnO 2H O 2e MnO 4OH 4 2 2        On addition, we get : 4 4 2 4 2 2MnO MnO 2H O 2MnO MnO 4OH          4 2 4 2 3MnO 2H O 2MnO MnO 4OH        4 4 2 1 MnO MnO 3    ; 4 2 1 1 MnO MnO 3   8.(A) 2 2 2 2 3 2 3 3 2 6 3 RH Ca RCa 2H , pH 2, H 10 M [Ca ] 5 10 , mass of Ca 40 5 10 g 40 5 10 Hardness in ppm of Ca 10 200 10                          9.(D) Initial mmol of KOH = 500 × 0.1 = 50 2 2 3 2 2KOH CO K CO H O    K CO BaCl BaCO 2KCl 2 3 2 3    KOH HCl KCl H O    2 mmol of HCl = 30 × 0.1 = 3 ; mmol of KOH left unreacted in 50 mL solution = (50×0.1) –3 = 2 millimoles of KOH in 50 mL reacted with CO2 = 2 ; millimoles of KOH in 500 mL reacted with CO2 = 20 2 mol KOH = 1 mol CO2 ; 20 mmol KOH = 10 mmol CO2 Volume of CO2 = 10 × 10–3 × 22.4 L ppm of CO2 = 3 10 10 22.4 6 10 1000 224      ppm 10.(C) mmol of BaCO3 = mmol of K2CO3 = mmol of CO2 = 1(considering in 50 mL) 3 3 BaCO BaCO m 197 1000 1, m 0.197 g 197 1000     11.(A) 2 4 3 4 1 mol 3 mol Fe Fe (SO ) 3FeSO   2 4 3 4 2 4 2 2 4 3 2 4 3 4 2 4 2 2 4 3 4 4 1mol Fe (SO ) .(NH ) SO . 24H O 1mole Fe (SO ) 2.41 2.41 3 2.41 molFe (SO ) .(NH ) SO . 24H O mole Fe (SO ) mol FeSO 964 964 964 moles of FeSO 0.0075     
Vidyamandir Classes APP | Stoichiometry-I & II 3 Solutions | Chemistry 12.(B) 2 4 3 4 4 1 mol 2 mol Cu Fe (SO ) 2FeSO CuSO    4 2 2 7 4 2 4 3 4 2 4 2 2 4 3 4 4 [Note : only FeSO reacts with K Cr O and not CuSO as Cu is in its highest O.S ]. 2.41 2.41 2 2.41 molFe (SO ) .(NH ) SO . 24H O mole Fe (SO ) mol FeSO 964 964 964 moles of FeSO 0.005      13.(C) meq of FeSO4 = meq of K2Cr2O7 3 2.41 1 3 1 10 6 V 964 60             ( n-factor of FeSO 1 4  and K Cr O 6 2 2 7   )   V 75 mL 14.(AC) Cu2+ does not react with MnO 4  . Only 2 C O2 4  will react 2 4 2 4 2 2 4 meq of MnO meq of C O 1 20 5 meq of C O 25 4         2 2 4 f m eq 25 m mol of C O 12.5 n 2     f f f f 2 2 2 2 n 1 n 1/2 2 2 2 2 3 4 6 n 2 n 1 2Cu 4I Cu I I I 2S O 2I S O                meq of Cu2+ = meq of KI = meq of I2 = meq of 2 S O2 3  = 1 25 1 10   = 2.5 2 2.5 m mol of Cu (n 1) 2.5 1      Difference in mmol of 2 C O2 4  and Cu2+ = 12.5–2.5 = 10 15.(ABCD) 7 2 4 2 Ca Mn O Mn n 10           meq of Ca(MnO4)2 = nMV = 10 × 1 100 100 meq 10   meq of FeSO n 1 1 1 100 100 meq 4        ; 2 4 100 meq of FeC O (n 3) 3 1 100 meq 3      meq of K Cr O (n 6) 6 1 16.6 100 meq 2 2 7      ; 2 meq of C O (n 2) 2 1 50 100 meq 2 4       16.(ABC) 2 Na HPO 2Na HPO 2 4 4      2 4 2 4 3 4 acid Base (n 1) n 1 HPO H H PO H H PO             2 4 3 4 (n 2) base HPO 2H H PO       ; 2 3 4 4 Acid (n 1) HPO H PO        17.(BCD) meq of H O meq of KMnO 2 2 4 w 1 1000 1 100 w 0.34 g 34 / 2 5       
Vidyamandir Classes APP | Stoichiometry-I & II 4 Solutions | Chemistry 2 4 4 2 2 2 In basic medium : MnO MnO (n 1) H O O (n 2)       18.(AB) 2 2 2 120 mmol 100 mmol 2HCl Ba(OH) BaCl 2H O    L.R is HCl (Also 20 mL of water is also added) mmol Ba(OH)2 left = (100–60) = 40 mmol of OH– in excess = 40 × 2 = 80 80 [OH ] 0.8 100    ; mmol of BaCl2 = 60 ; mmol of Cl– = 120 [Cl– ] = 120 1.2M 100  ; mmol of Ba(OH)2 left + mmol of BaCl2 = 40 + 60 = 100 2 100 [Ba ] 1 100    19.(ABD) H O2 + SO3  H SO 2 4 18 80 98 12 80 12 18  98 12 18  = 53.3 g = 65.33 g (100 – 53.3) = 46.7 g of H2SO4 is initially present and 65.33 g of H2SO4 is newly formed on add of 12 g of water. H2SO4 total = 65.33 + 46.7 = 112 g But on addition of 9 g of water, moles of free 3 12 9 1 SO 18 18 6    Addition of 9 g of water will produce 98 9 49 g 18   of newly formed H2SO4. Total H2SO4 = 49 + 46.7 = 95.7 20.(BD) 2 mol of HCO3  will make one mole each of CaCO3, CaCl2 and MgCl2 to have equal hardness ppm of 6 HCO 61 2 122g in 10 mL of H O 3 2     3 1 mol of CaCO 100 ppm  ; 2 1 mol of CaCl 111 ppm  ; 2 1 mol of MgCl 95 ppm  21.(AB) H2O is L.R. 6H O 4Fe(OH) 2 3 4 2 mole 2 1.33 6    22.(AB) Volume strength = 5.6 × N NfVf = N1V1 + N2V2 + N3V3 0.5 L of each solution and 1.5 L of water makes final solution 3L final f 10 15 20 N 3 0.5 0.5 0.5 5.6 5.6 5.6 N 1.34         Volume strength of final solution = 1.34 ×5.6 = 7.5